The second Hankel determinant for logarithmic coefficients of inverse functions of bounded turning of a given order

Abstract

The sharp bound of the second Hankel determinant of logarithmic coefficients of inverse functions of bounded turning of order $\alpha$ is computed.

1 Introduction

For $r>0,$ let ${\mathbb {D}}_r:=\{ z\in {\mathbb {C}}: |z|<r \}$, ${\mathbb {D}}:={\mathbb {D}}_1$, $\overline{{\mathbb {D}}}:=\{z\in {\mathbb {C}}:|z|\le 1\}$ and let $\mathbb T:=\{z\in {\mathbb {C}}:|z|=1\}.$ Let $\mathcal {H}({\mathbb {D}}_r)$ denote the class of all analytic functions f in ${\mathbb {D}}_r$ and let $\mathcal H:=\mathcal H(\mathbb D).$ Then $f\in \mathcal {H}({\mathbb {D}}_r)$ has the following representation

$$\begin{aligned} f(z) = \sum _{n=0}^{\infty }a_n(f) z^n,\quad z\in {\mathbb {D}}_r. \end{aligned}$$

(1.1)

Let ${\mathcal {A}}({\mathbb {D}}_r)$ be the subclass of ${\mathcal H}({\mathbb {D}}_r)$ of all f normalized by $f(0)=0=f'(0)-1$ and let $\mathcal {A}:=\mathcal {A}({\mathbb {D}}).$ By $\mathcal {S}$ we denote the subclass of all univalent (i.e., analytic and injective in $\mathbb D$) functions in $\mathcal {A}$.

Given $\alpha \in [0,1),$ by $\mathcal {P}'(\alpha )$ we denote the class of all functions $f\in {\mathcal {A}}$ such that

$$\begin{aligned} {{\,\textrm{Re}\,}}f'(z) >\alpha ,\quad z\in {\mathbb {D}}. \end{aligned}$$

(1.2)

Such functions are called of bounded turning of order $\alpha .$ Particularly, elements of $\mathcal P':=\mathcal P'(0)$ are called of bounded turning (cf. [7, Vol. I, p. 101]). Recall also that the condition (1.2) with $\alpha =0$ is known as a famous criterium of univalence due to Alexander [1] (cf. [7, Vol. I, Theorem 12, p. 88]) which means that $\mathcal P'\subset \mathcal S.$ Since $\mathcal P'(\alpha )\subset \mathcal P'$ for $\alpha \in [0,1),$ we see that $\mathcal P'(\alpha )\subset \mathcal S$ for every $\alpha \in [0,1).$ The class $\mathcal P'$ is one of the fundamental subfamily of univalent functions and has been extensively studied by many authors e.g., [15, 16].

If $f\in \mathcal S,$ then $f^{-1}\in \mathcal {H}\left( \mathbb D_{r(f)}\right)$, where $r(f):=\sup (\{r>0:\mathbb D_r\subset f(\mathbb D)\}).$ Thus

$$\begin{aligned} f^{-1}(w)=w+\sum _{n=2}^\infty A_nw^n,\quad w\in \mathbb D_{r(f)}, \end{aligned}$$

(1.3)

where $A_n:=a_n(f^{-1}).$ By Koebe One-Quarter Theorem (e.g., [5, p. 31]), it follows that $r(f) \ge 1/4$ for every $f\in \mathcal S.$

For $f\in \mathcal S$ define

$$\begin{aligned} L_f(z):= \frac{1}{2}\log \frac{f(z)}{z}=\sum _{n=1}^\infty \gamma _n z^n,\quad z\in \mathbb D, \end{aligned}$$

a logarithmic function associated with f. The numbers $\gamma _n:=a_n(L_f)$ are called the logarithmic coefficients of f. It is well-known, that the logarithmic coefficients play a crucial role in Milin’s conjecture (see [17, 5, p. 155]).

Referring to the above idea, for $f\in \mathcal S$ there exists the unique function $L_{f^{-1}}\in \mathcal {H}\left( \mathbb D_{r(f)}\right)$ such that

$$\begin{aligned} L_{f^{-1}}(w):=\frac{1}{2}\log \frac{f^{-1}(w)}{w}=\sum _{n=1}^\infty \varGamma _n w^n,\quad w\in {\mathbb {D}}_{r(f)}, \end{aligned}$$

(1.4)

where $\varGamma _n:=a_n\left( L_{f^{-1}}\right)$ are logarithmic coefficients of the inverse function $f^{-1}.$

From (1.3), it follows that (e.g., [7, Vol. I, p. 57])

$$\begin{aligned} A_2=-a_2,\quad A_3=-a_3+2a_2^2,\quad A_4=-a_4+5a_2a_3-5a_2^3, \end{aligned}$$

(1.5)

where $a_n:=a_n(f).$ Thus, from (1.4), we derive that

$$\begin{aligned} \varGamma _1=\frac{1}{2}A_2,\quad \varGamma _2=\frac{1}{2}A_3-\frac{1}{4}A_2^2,\quad \varGamma _3=\frac{1}{2}A_4-\frac{1}{2}A_2A_3+\frac{1}{6}A_2^3, \end{aligned}$$

and next using (1.5), we obtain

$$\begin{aligned} \begin{aligned}&\varGamma _1=-\frac{1}{2}a_2,\quad \varGamma _2=-\frac{1}{2}a_3+\frac{3}{4}a_2^2,\quad \varGamma _3=-\frac{1}{2}a_4+2a_2a_3-\frac{5}{3}a_2^3. \end{aligned} \end{aligned}$$

(1.6)

For $q,n\in \mathbb N,$ the Hankel matrix $H_{q,n}(f)$ of $f\in \mathcal A$ of the form (1.1) is defined as

$$\begin{aligned} H_{q,n}(f):= \left[ \begin{matrix} a_{n} &{} a_{n+1}&{} \cdots &{} a_{n+q-1} \\ a_{n+1} &{} a_{n+2} &{} \cdots &{} a_{n+q}\\ \vdots &{} \vdots &{} \vdots &{} \vdots \\ a_{n+q-1} &{} a_{n+q} &{}\cdots &{} a_{n+2(q-1)} \end{matrix}\right] . \end{aligned}$$

(1.7)

In recent years, there has been a great deal of attention devoted to finding bounds for the modulus of the second and third Hankel determinants $\det H_{2,2}(f)$ and $\det H_{3,1}(f)$, when f belongs to various subclasses of ${\mathcal {A}}$ (see [2, 11, 12] for further references).

Based on these ideas, in [9, 10], the authors started the study the Hankel determinant $\det H_{q,n}(L_f)$ whose entries are logarithmic coefficients of $f\in \mathcal S,$ that is, $a_n$ in (1.7) are replaced by $\gamma _n.$ In this paper, we continue analogous research considering the Hankel determinant $\det H_{q,n}(L_{f^{-1}})$ whose entries are logarithmic coefficients of inverse functions, i.e., $a_n$ in (1.7) are now replaced by $\varGamma _n.$ Such research can be found in [6, 14]. This paper demonstrates the sharp estimate of

$$\begin{aligned} \left| \det H_{2,1}\left( L_{f^{-1}}\right) \right| =\left| \varGamma _1\varGamma _3-\varGamma _2^2\right| =\frac{1}{48}\left| 13a_2^4-12a_3^2+12a_2a_4-12a_2^2a_3\right| \end{aligned}$$

in the class $\mathcal P'(\alpha ).$

2 Preliminary lemmas

Denote by $\mathcal {P}$ the class of analytic functions $p\in \mathcal H$ with positive real part given by

$$\begin{aligned} p(z)=1+\sum _{n=1}^{\infty }c_n z^n,\quad z\in {\mathbb {D}}, \end{aligned}$$

(2.1)

where $c_n:=a_n(p).$

In the proof of the main result, we will use the following lemma which contains the well-known formula for $c_2$ (see e.g., [18, p. 166]) and the formula for $c_3$ (see [3, Lemma 2.4] with further remarks related to extremal functions).

Lemma 2.1

If $p \in \mathcal {P}$ is of the form (2.1), then

$$\begin{aligned} c_1= & {} 2\zeta _1, \end{aligned}$$

(2.2)

$$\begin{aligned} c_2= & {} 2\zeta _1^2 + 2(1-|\zeta _1|^2)\zeta _2 \end{aligned}$$

(2.3)

and

$$\begin{aligned} c_3 = 2\zeta _1^3+2(1-|\zeta _1|^2)(2\zeta _1-\overline{\zeta _1}\zeta _2)\zeta _2 + 2(1-|\zeta _1|^2)(1-|\zeta _2|^2)\zeta _3 \end{aligned}$$

(2.4)

for some $\zeta _1,\zeta _2, \zeta _3 \in \overline{{\mathbb {D}}}.$

For $\zeta _1 \in {\mathbb {T}}$, there is a unique function $p \in \mathcal {P}$ with $c_1$ as in (2.2), namely

$$\begin{aligned} p(z) = \frac{1+\zeta _1 z}{1-\zeta _1 z}, \quad z\in {\mathbb {D}}. \end{aligned}$$

For $\zeta _1\in {\mathbb {D}}$ and $\zeta _2 \in {\mathbb {T}}$, there is a unique function $p \in \mathcal {P}$ with $c_1$ and $c_2$ as in (2.2) and (2.3), namely,

$$\begin{aligned} p(z) = \frac{1+( {\overline{\zeta }}_1 \zeta _2 +\zeta _1 )z + \zeta _2 z^2}{1+( {\overline{\zeta }}_1 \zeta _2 -\zeta _1 )z - \zeta _2 z^2}, \quad z\in {\mathbb {D}}. \end{aligned}$$

(2.5)

Lemma 2.2

[4] For real numbers A, B, C, let

$$\begin{aligned} Y(A,B,C):= \max \left( \left\{ |A+Bz+Cz^2|+1-|z|^2: z\in \overline{{\mathbb {D}}}\right\} \right) . \end{aligned}$$

I. If $AC\ge 0,$ then

$$\begin{aligned} Y(A,B,C)=\left\{ \begin{array}{ll} |A|+|B|+|C|, &{} |B|\ge 2(1-|C|),\\ 1+|A|+\dfrac{B^2}{4(1-|C|)}, &{} |B|<2(1-|C|). \end{array} \right. \end{aligned}$$

II. If $AC<0,$ then

$$\begin{aligned} Y(A,B,C)=\left\{ \begin{array}{lll} 1-|A|+\dfrac{B^2}{4(1-|C|)}, &{} -4AC(C^{-2}-1)\le B^2 \wedge |B|<2(1-|C|), \\ 1+|A|+\dfrac{B^2}{4(1+|C|)}, &{} B^2<\min \left\{ 4(1+|C|)^2,-4AC(C^{-2}-1)\right\} , \\ R(A,B,C), &{} \textrm{otherwise}, \end{array} \right. \end{aligned}$$

where

$$\begin{aligned} R(A,B,C):=\left\{ \begin{array}{lll} |A|+|B|-|C|, &{} |C|(|B|+4|A|)\le |AB|,\\ -|A|+|B|+|C|, &{} |AB|\le |C|(|B|-4|A|), \\ (|C|+|A|)\sqrt{1-\dfrac{B^2}{4AC}}, &{} \textrm{otherwise}. \end{array} \right. \end{aligned}$$

We recall now Laguerre’s rule of counting zeros of polynomials in an interval (see [8, 13, 19, pp. 19–20]), which we apply in the proof of the main theorem. Given a real polynomial

$$\begin{aligned} Q(u):= d_0u^n + d_1u^{n-1} + \cdots + d_{n-1}u + d_n,\quad u\in \mathbb R,\ d_0,\dots ,d_n\in \mathbb R, \end{aligned}$$

consider a finite sequence $(q_k), k = 0, 1,\ldots , n,$ of polynomials of the form

$$\begin{aligned} q_k(u):=\sum _{j=0}^k d_ju^{k-j},\quad u\in \mathbb R. \end{aligned}$$

For each $u_0\in {\mathbb {R}},$ let $N(Q; u_0)$ denote the number of sign changes in the sequence $(q_k(u_0)), k = 0, 1,\ldots , n.$ Given an interval $I \subset {\mathbb {R}},$ denote by Z(Q; I) the number of zeros of Q in I counted with their orders. Then the following theorem due to Laguerre holds.

Theorem 2.1

If $a<b$ and $Q(a)Q(b)\ne 0,$ then

$$\begin{aligned} Z(Q; (a, b)) = N(Q; a) - N(Q; b) \end{aligned}$$

or

$$\begin{aligned} N(Q; a) - N(Q; b) - Z(Q; (a, b)) \end{aligned}$$

is an even positive integer.

Note that

$$\begin{aligned} q_k(0) = d_k,\quad q_k(1) =\sum _{j=0}^k d_j. \end{aligned}$$

Thus, when $[a, b]:= [0, 1],$ Theorem 2.1 reduces to the following useful corollary.

Corollary 2.1

If $Q(0)Q(1)\ne 0,$ then

$$\begin{aligned} Z(Q; (0, 1)) = N(Q; 0) - N(Q; 1) \end{aligned}$$

or

$$\begin{aligned} N(Q; 0) - N(Q; 1) - Z(Q; (0, 1)) \end{aligned}$$

is an even positive integer, where N(Q; 0) and N(Q; 1) are the numbers of sign changes in the sequence of polynomial coefficients $(d_k)$ and in the sequence of sums $(\sum _{j=0}^k d_j ),$ where $k = 0, 1,\ldots , n,$ respectively.

3 Main result

Now we prove the main theorem of this paper.

Theorem 3.1

Let $\alpha \in [0,1).$ If $f\in {{\mathcal {P}}'}(\alpha ),$ then

$$\begin{aligned} \left| \varGamma _1\varGamma _3-\varGamma _2^2\right| \le {\left\{ \begin{array}{ll} &{}\dfrac{1}{144}(\alpha -1)^2(39\alpha ^2-54\alpha +17),\quad 0\le \alpha<\alpha _0,\\ &{} \dfrac{1}{1872}(256\alpha ^2-440\alpha +211),\quad \alpha _0\le \alpha <\dfrac{1}{4},\\ &{} \dfrac{1}{9} (1-\alpha )^2,\quad \dfrac{1}{4}\le \alpha \le 1,\\ \end{array}\right. } \end{aligned}$$

(3.1)

where $\alpha _0\approx 0.014779$ is a unique root in [0, 1) of the equation

$$\begin{aligned} 507\alpha ^4 - 1716\alpha ^3 + 1876\alpha ^2 - 704\alpha + 10=0. \end{aligned}$$

All inequalities are sharp.

Proof

Let $\alpha \in [0,1)$ and $f\in {{\mathcal {P}}}'(\alpha )$ be of the form (1.1). Then by (1.2), there exists $p\in {\mathcal {P}}$ of the form (2.1) such that

$$\begin{aligned} f'(z)=(1-\alpha )p(z)+\alpha ,\quad z\in {\mathbb {D}}. \end{aligned}$$

(3.2)

Putting the series (1.1) and (2.1) into (3.2), by equating the coefficients, we get

$$\begin{aligned} a_2(f)=\frac{1}{2}(1-\alpha )c_1,\quad a_3(f)=\frac{1}{3}(1-\alpha )c_2,\quad a_4(f)=\frac{1}{4}(1-\alpha )c_3. \end{aligned}$$

(3.3)

Hence and from (1.6), we obtain

$$\begin{aligned} \begin{aligned}&\varGamma _1 =-\frac{1}{4}(1-\alpha )c_1, \quad \varGamma _2 =-\frac{1}{48}(1-\alpha )\left( 8c_2-9(1-\alpha )c_1^2\right) ,\\&\varGamma _3 =-\frac{1}{24}(1-\alpha )\left( 3c_3-8(1-\alpha )c_1c_2+5(1-\alpha )^2c_1^3\right) \end{aligned} \end{aligned}$$

(3.4)

and therefore

$$\begin{aligned} \begin{aligned}&\varGamma _1\varGamma _3-\varGamma _2^2\\&\quad =\frac{1}{2304}(1-\alpha )^2\left( 39(1-\alpha )^2c_1^4-48(1-\alpha )c_1^2c_2+72c_1c_3-64c_2^2\right) . \end{aligned} \end{aligned}$$

(3.5)

Since the class ${{\mathcal {P}}}'(\alpha )$ and $|H_{2,1}\left( F_{f^{-1}}\right) |$ are rotationally invariant, without loss of generality we may assume that $a_2\ge 0,$ which in view of (3.3) yields $c_1 \in [0,2],$ i.e., by (2.2) that $\zeta _1\in [0,1].$ Thus, substituting (2.2)–(2.4) into (3.5), we obtain

$$\begin{aligned} \begin{aligned} \varGamma _1\varGamma _3-\varGamma _2^2 =&\frac{1}{144}(1-\alpha )^2\left( (39\alpha ^2-54\alpha +17)\zeta _1^4+4(6\alpha -5)(1-\zeta _1^2)\zeta _1^2\zeta _2\right. \\&\left. -2(1-\zeta _1^2)(\zeta _1^2+8)\zeta _2^2 +18\zeta _1(1-\zeta _1^2)(1-|\zeta _2|^2)\zeta _3\right) \\ \end{aligned} \end{aligned}$$

(3.6)

for some $\zeta _1\in [0,1]$ and $\zeta _2,\zeta _3 \in \overline{{\mathbb {D}}}$.

A. Suppose that $\zeta _1=0.$ Then from (3.6),

$$\begin{aligned} \left| \varGamma _1\varGamma _3-\varGamma _2^2\right| =\frac{1}{9}(1-\alpha )^2|\zeta _2|^2\le \frac{1}{9} (1-\alpha )^2. \end{aligned}$$

B. Suppose that $\zeta _1=1.$ Then from (3.6),

$$\begin{aligned} \left| \varGamma _1\varGamma _3-\varGamma _2^2\right| =\frac{1}{144}(1-\alpha )^2(39\alpha ^2-54\alpha +17). \end{aligned}$$

C. Suppose that $\zeta _1\in (0,1).$ Since $\zeta _3 \in \overline{{\mathbb {D}}}$, from (3.5) we get

$$\begin{aligned} \left| \varGamma _1\varGamma _3-(\varGamma _2)^2\right| \le \frac{1}{8}(1-\alpha )^2\zeta _1(1-\zeta _1^2) \varPhi (A,B,C), \end{aligned}$$

where

$$\begin{aligned} \varPhi (A,B,C):= \left| A +B\zeta _2 +C\zeta _2^2 \right| + 1 -|\zeta _2|^2, \end{aligned}$$

with

$$\begin{aligned} A:=\frac{(39\alpha ^2-54\alpha +17)\zeta _1^3}{18(1-\zeta _1^2)},\quad B:=\frac{2(6\alpha -5)\zeta _1}{9},\quad C:=-\frac{\zeta _1^2+8}{9\zeta _1}. \end{aligned}$$

Observe that $A\le 0$ if and only if $\alpha \in [\alpha _1,\alpha _2],$ where

$$\begin{aligned} \alpha _1:=\frac{1}{39}(27-\sqrt{66})\approx 0.484,\quad \alpha _2:= \frac{1}{39}(27+\sqrt{66})\approx 0.901. \end{aligned}$$

For further argumentation, we apply Lemma 2.2.

C.I. Consider the case $AC\ge 0$ which holds if and only if $\alpha \in [\alpha _1,\alpha _2],$ i.e., Part I of Lemma 2.2. We show that this case reduces to the condition $|B|\ge 2(1-|C|)$ only.

(a) If $\alpha \in [\alpha _1,5/6],$ then the condition $|B|\ge 2(1-|C|)$ is equivalent to

$$\begin{aligned} \frac{2}{9\zeta _1}(6(\alpha -1)\zeta _1^2+9\zeta _1-8)=\frac{2}{9\zeta _1}\left( (6\alpha -5)\zeta _1^2-(1-\zeta _1)(8-\zeta _1)\right) \le 0, \end{aligned}$$

which clearly holds. Applying Lemma 2.2 for $0<\zeta _1<1$ and $\alpha \in [\alpha _1,5/6]$, we get

$$\begin{aligned} \begin{aligned} \left| \varGamma _1\varGamma _3-(\varGamma _2)^2\right|&\le \frac{1}{8}(1-\alpha )^2\zeta _1(1-\zeta _1^2)\varPhi (A,B,C)\\&\le \frac{1}{8}(1-\alpha )^2\zeta _1(1-\zeta _1^2)Y(A,B,C)\\&=\frac{1}{8}(1-\alpha )^2\zeta _1(1-\zeta _1^2)(|A|+|B|+|C|)=\rho (\zeta _1), \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \rho (t):=-\frac{1}{144}(\alpha -1)^2\left( 39(1-\alpha )^2t^4+6(4\alpha -1)t^2-16\right) ,\quad t\in [0,1]. \end{aligned}$$

Since the equation

$$\begin{aligned} \rho '(t)=-\frac{1}{12}(1-\alpha )^2\left( 13(1-\alpha )^2t^2+4\alpha -1\right) t=0,\quad t\in (0,1), \end{aligned}$$

has no root, it follows that $\rho$ is decreasing and therefore

$$\begin{aligned} \rho (t)\le \rho (0)=\frac{1}{9}(1-\alpha )^2,\quad t\in [0,1]. \end{aligned}$$

(b) If $\alpha \in (5/6,\alpha _2],$ then the condition $|B|\ge 2(1-|C|)$ is equivalent to

$$\begin{aligned} \frac{2}{9\zeta _1}((6\alpha -4)\zeta _1^2-9\zeta _1+8)=\frac{2}{9\zeta _1}((6\alpha -5)\zeta _1^2+(8-\zeta _1)(1-\zeta _1))\ge 0 \end{aligned}$$

which is true for $\zeta _1\in (0,1).$ Applying Lemma 2.2 for $\zeta _1\in (0,1)$ and $\alpha \in (5/6,\alpha _2)$, we get

$$\begin{aligned} \Big |\varGamma _1\varGamma _3-\varGamma _2^2\Big | \le \frac{1}{8}(1-\alpha )^2\zeta _1(1-\zeta _1^2)\left( |A|+|B|+|C| \right) =\varrho (\zeta _1), \end{aligned}$$

where

$$\begin{aligned} \varrho (t):=-\frac{1}{144}(1-\alpha )^2((39\alpha ^2-30\alpha -1)t^4-2(-17+12\alpha )t^2-16),\quad t\in [0,1]. \end{aligned}$$

Since the equation

$$\begin{aligned} \varrho '(t)=-\frac{1}{36}(1-\alpha )^2 t((39\alpha ^2-30\alpha -1)t^2+17-12\alpha ))=0,\quad t\in (0,1), \end{aligned}$$

has no root, it follows that $\varrho$ is decreasing and therefore

$$\begin{aligned} \varrho (t)\le \varrho (0)=\frac{1}{9}(1-\alpha )^2,\quad t\in [0,1]. \end{aligned}$$

C.II. Now we consider the case $AC< 0$ which holds if and only if $\alpha \in [0,\alpha _1)\cup (\alpha _2,1).$

C.II.1. Let’s consider the condition $|B|< 2(1-|C|).$

(a) If $\alpha \in [0,\alpha _1),$ then the condition $|B|< 2(1-|C|)$ is equivalent to

$$\begin{aligned} \frac{2}{9\zeta _1}(6(\alpha -1)\zeta _1^2+9\zeta _1-8)=\frac{2}{9\zeta _1}\left( (6\alpha -5)\zeta _1^2-(1-\zeta _1)(8-\zeta _1)\right) < 0, \end{aligned}$$

which is false for $\zeta _1\in (0,1).$

(b) If $\alpha \in (\alpha _2,1),$ then the condition $|B|< 2(1-|C|)$ is equivalent to

$$\begin{aligned} \frac{2}{9\zeta _1}((6\alpha -4)\zeta _1^2-9\zeta _1+8)=\frac{2}{9\zeta _1}((6\alpha -5)\zeta _1^2+(8-\zeta _1)(1-\zeta _1))< 0 \end{aligned}$$

which is false for $\zeta _1\in (0,1).$

C.II.2. Since

$$\begin{aligned} -4AC\left( \frac{1}{C^2}-1\right) =-\frac{2(39\alpha ^2-54\alpha +17)\zeta _1^2(64-\zeta _1^2)}{81(\zeta _1^2+8)} \end{aligned}$$

and

$$\begin{aligned} 4(1+|C|)^2=\frac{4(\zeta _1+8)^2(\zeta _1+1)^2}{81\zeta _1^2}, \end{aligned}$$

it follows that the condition $B^2<\min \{4(1+|C|)^2,-4AC(C^{-2}-1)\}$ is equivalent to

$$\begin{aligned} \frac{2\zeta _1^2}{27{\zeta _1^2+8}}\left( 11\zeta _1^2(\alpha -1)^2+1024\alpha ^2-1472\alpha +496\right) <0. \end{aligned}$$

(3.7)

However

$$\begin{aligned} 11\zeta _1^2(\alpha -1)^2+1024\alpha ^2-1472\alpha +496\ge 1024\alpha ^2-1472\alpha +496\ge 0 \end{aligned}$$

for $\alpha \in (0,(23-\sqrt{33})/32]\cup [(23+\sqrt{33})/32],1).$ But $\alpha _1<(23-\sqrt{33})/32$ and $\alpha _2>(23+\sqrt{33})/32$ which yields that the inequality (3.7) is false for $\zeta _1\in (0,1)$ and $\alpha \in [0,\alpha _1)\cup (\alpha _2,1).$

C.II.3. Now we consider the condition $|C|(|B|+4|A|)\le |AB|.$

(a) Suppose that $\alpha \in [0,\alpha _1).$ Then the condition $|C|(|B|+4|A|)\le |AB|$ is equivalent to

$$\begin{aligned} \begin{aligned}&\frac{1}{81(1-\zeta _1^2)}\left( 234\alpha ^3-441\alpha ^2+276\alpha -61)\zeta _1^4\right. \\&\quad \left. +2(312\alpha ^2-390\alpha +101)\zeta _1^2-96\alpha +80\right) \le 0, \end{aligned} \end{aligned}$$

(3.8)

which is equivalent to

$$\begin{aligned} \varphi (\zeta _1^2)\le 0, \end{aligned}$$

(3.9)

where for $t\in \mathbb R,$

$$\begin{aligned} \varphi (t):=(234\alpha ^3-441\alpha ^2+276\alpha -61)t^2 +2t(312\alpha ^2-390\alpha +101)-96\alpha +80. \nonumber \\ \end{aligned}$$

(3.10)

Observe that for $\alpha \in [0,\alpha _1)$ the inequalities $234\alpha ^3-441\alpha ^2+276\alpha -61<0$ and $\Delta :=479232\alpha ^4-1217664\alpha ^3+1107600\alpha ^2-426864\alpha +60324>0$ are true because the discriminant $\Delta$ has two roots in [0, 1) namely, $\alpha '\approx 0.87646$ and $\alpha ''\approx 0.88376.$ Hence the square trinomial $\varphi$ has two roots

$$\begin{aligned} \begin{aligned} t_{1,2}&:=\frac{1}{{2(234\alpha ^3-441\alpha ^2+276\alpha -61)}}\left( -2(312\alpha ^2-390\alpha +101)\right. \\&\quad \left. \pm \sqrt{479232\alpha ^4-1217664\alpha ^3+1107600\alpha ^2-426864\alpha +60324}\right) . \end{aligned} \end{aligned}$$

(3.11)

We will show that $t_1<0,$ i.e., equivalently that

$$\begin{aligned} \begin{aligned}&\sqrt{479232\alpha ^4-1217664\alpha ^3+1107600\alpha ^2-426864\alpha +60324}\\&\quad >2(312\alpha ^2-390\alpha +101). \end{aligned} \end{aligned}$$

(3.12)

If $\alpha \in [\alpha _9,\alpha _1),$ where $\alpha _9:=(5 - \sqrt{167/39})/8\approx 0.36634,$ then the inequality (3.12) is obviously true. For $\alpha \in [0,\alpha _9)$, by squaring both sides of the inequality (3.12), we get the inequality

$$\begin{aligned} 64(6\alpha -5) (234\alpha ^3 - 441\alpha ^2 + 276\alpha - 61)>0 \end{aligned}$$

which is true.

Moreover, the inequality $t_2>1$ is equivalent to the inequality

$$\begin{aligned} \begin{aligned}&-\sqrt{479232\alpha ^4-1217664\alpha ^3+1107600\alpha ^2-426864\alpha +60324}\\&\quad < 468\alpha ^3-258\alpha ^2-228\alpha +80, \end{aligned} \end{aligned}$$

(3.13)

which is obviously true for $\alpha \in [0,\alpha _3],$ where $\alpha _3\approx 0.303972$ is the smallest positive root of the equation $468\alpha ^3-258\alpha ^2-228\alpha +80=0$. For $\alpha \in (\alpha _3,\alpha _1)$, by squaring both sides of the inequality (3.13) and grou**, we get the true inequality

$$\begin{aligned} 4(6\alpha +13)(234\alpha ^3-441\alpha ^2+276\alpha -61)(39\alpha ^2-54\alpha +17)< 0. \end{aligned}$$

Thus, we conclude that for $\alpha \in [0,\alpha _1)$ the inequality (3.9) is false.

(b) If $\alpha \in (\alpha _2,1),$ then the condition $|C|(|B|+4|A|)\le |AB|$ is equivalent to

$$\begin{aligned} \frac{3\zeta _1^4(78\alpha -43)(\alpha -1)^2-6\zeta _1^2(104\alpha ^2-158\alpha +57)-96\alpha +80}{81(1-\zeta _1^2)}\ge 0. \end{aligned}$$

(3.14)

which is equivalent to

$$\begin{aligned} \phi (\zeta _1^2)\ge 0, \end{aligned}$$

(3.15)

where for $t\in \mathbb R,$

$$\begin{aligned} \phi (t):=3(78\alpha -43)(\alpha -1)^2t^2-6(104\alpha ^2-158\alpha +57)t-96\alpha +80. \end{aligned}$$

(3.16)

Note that the inequalities $3(78\alpha -43)(\alpha -1)^2>0$ and $\Delta :=479232\alpha ^4-1487232\alpha ^3+1705488\alpha ^2-855408\alpha +158244>0$ are true for $\alpha \in (\alpha _2,1)$. Hence, the square trinomial $\phi$ has two roots

$$\begin{aligned} \begin{aligned} t_{3,4}&:=\frac{1}{{6(78\alpha -43)(\alpha -1)^2}}\left( 6(104\alpha ^2-158\alpha +57)\right. \\&\quad \left. \pm \sqrt{479232\alpha ^4-1487232\alpha ^3+1705488\alpha ^2-855408\alpha +158244}\right) . \end{aligned} \end{aligned}$$

(3.17)

We will show that $t_4<0,$ i.e., equivalently that

$$\begin{aligned} \begin{aligned}&\sqrt{479232\alpha ^4-1487232\alpha ^3+1705488\alpha ^2-855408\alpha +158244}\\&\quad >6(104\alpha ^2-158\alpha +57). \end{aligned} \end{aligned}$$

(3.18)

If $\alpha \in (\alpha _2,\alpha _{10}],$ where $\alpha _{10}:=(79+\sqrt{313})/104\approx 0.929729,$ then the inequality (3.18) is obviously true. For $\alpha \in (\alpha _{10},1)$ by squaring both sides of the inequality (3.18) and grou**, we get the inequality

$$\begin{aligned} 192(78\alpha -43)(6\alpha -5)(\alpha -1)^2>0 \end{aligned}$$

which is true.

Moreover, the inequality $t_1>1$ is equivalent to the inequality

$$\begin{aligned} \begin{aligned}&\sqrt{479232\alpha ^4-1487232\alpha ^3+1705488\alpha ^2-855408\alpha +158244}\\&\quad > 468\alpha ^3-1818\alpha ^2+1932\alpha -600, \end{aligned} \end{aligned}$$

(3.19)

which is obviously true for $\alpha \in [\alpha _{12},1),$ where $\alpha _{12}\approx 0.933423$ is the root of the equation $468\alpha ^3-1818\alpha ^2+1932\alpha -600=0.$ For $\alpha \in (\alpha _2,\alpha _{12})$, by squaring both sides of the inequality (3.19) and grou** we get the inequality

$$\begin{aligned} 12(78\alpha -43)(6\alpha -23)(39\alpha ^2-54\alpha +17)(\alpha -1)^2<0 \end{aligned}$$

which is true.

Thus, we conclude that for $\alpha \in (\alpha _2,1)$ the inequality (3.15) is false.

CII.4 Let’s consider the condition $|C|(|B|-4|A|)\ge |AB|.$

(a) If $\alpha \in (0,\alpha _1),$ then the condition $|C|(|B|-4|A|)\ge |AB|$ is equivalent to the inequality (3.14), which is equivalent to the inequality (3.15) with $\phi$ given by (3.16). We have $3(78\alpha -43)(\alpha -1)^2<0,$ $6(104\alpha ^2-158\alpha +57)>0$ and $\Delta :=479232\alpha ^4-1487232\alpha ^3+1705488\alpha ^2-855408\alpha +158244>0$ for $\alpha \in (0,\alpha _1)$. Hence the square trinomial $\varphi$ has two roots $t_{3,4}$ given by (3.17). Note that $t_3<0$ evidently. Moreover, the inequality $t_4>0$ is equivalent to

$$\begin{aligned} \begin{aligned}&6(104\alpha ^2-158\alpha +57)\\&\quad <\sqrt{479232\alpha ^4-1487232\alpha ^3+1705488\alpha ^2 -855408\alpha +158244}. \end{aligned} \end{aligned}$$

(3.20)

Squared on both sides inequality (3.20) and transferred to one side, we get the inequality

$$\begin{aligned} 192(78\alpha -43)(6\alpha -5)(\alpha -1)^2>0, \end{aligned}$$

which is true for $\alpha \in (0,\alpha _1).$ On the other hand the inequality $t_4<1$ is equivalent to

$$\begin{aligned} \begin{aligned}&-\sqrt{479232\alpha ^4-1487232\alpha ^3+1705488\alpha ^2-855408\alpha +158244}\\&\quad >468\alpha ^3-1818\alpha ^2+1932\alpha -600, \end{aligned} \end{aligned}$$

which after squaring both sides and grou** is equivalent to the inequality

$$\begin{aligned} 12(78\alpha -43)(6\alpha -23)(39\alpha ^2-54\alpha +17)(\alpha -1)^2>0 \end{aligned}$$

being true for $\alpha \in (0,\alpha _1).$ Thus, we conclude that for $\alpha \in (0,\alpha _1)$ the inequality (3.15) is true for $0<\zeta _1\le \zeta _1'$, where $\zeta _1':=\sqrt{t_4}.$ Applying Lemma 2.2 for $0<\zeta _1\le \zeta _1'$, we get

$$\begin{aligned} |\Gamma _1\Gamma _3-\Gamma _2^2| \le \frac{1}{8}\zeta _1(1-\zeta _1^2)\left( -|A|+|B|+|C|\right) =:\sigma (\zeta _1), \end{aligned}$$

where

$$\begin{aligned} \sigma (t):=-\frac{1}{144}(1-\alpha )^2\left( 39(\alpha -1)^2t^4+6(4\alpha -1)t^2-16\right) ,\quad t\in \mathbb R. \end{aligned}$$

We have

$$\begin{aligned} \sigma (0)=\frac{1}{9}(1-\alpha )^2 \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \sigma (\zeta _1')&=\frac{2}{9(78\alpha -43)^2}(-32448\alpha ^4+95082\alpha ^3-103517\alpha ^2+49654\alpha -8852\\&\quad \left. +(104\alpha ^2-144\alpha +149)\sqrt{119808\alpha ^4-371808\alpha ^3+426372\alpha ^2-213852\alpha +39561}\right) . \end{aligned} \end{aligned}$$

Differentiating $\sigma$ lead to the equation

$$\begin{aligned} \sigma '(t)=-\frac{1}{12}(1-\alpha )^2\left( 13(\alpha -1)^2t^2+4\alpha -1\right) t=0 \end{aligned}$$

For $\alpha \in (0,1/4]$, we have

$$\begin{aligned} \sigma (\zeta _1)\le \sigma (t_5)=\frac{1}{1872}(256\alpha ^2-440\alpha +211),\quad 0<\zeta _1\le \zeta _1', \end{aligned}$$

where $t_5:=\sqrt{13-52\alpha }/(13(1-\alpha )).$ For $\alpha \in (1/4,\alpha _1)$, we have

$$\begin{aligned} \sigma (\zeta _1)\le \sigma (0)=\frac{1}{9}(1-\alpha )^2,\quad 0<\zeta _1\le \zeta _1'. \end{aligned}$$

(b) If $\alpha \in (\alpha _2,1),$ then the condition $|C|(|B|-4|A|)\ge |AB|$ is equivalent to the inequality (3.8) which is equivalent to the inequality (3.9), where $\varphi$ is defined by (3.10). Observe that the inequalities $234\alpha ^3-441\alpha ^2+276\alpha -61>0,$ $2(312\alpha ^2-390\alpha +101)>0$ and $\Delta :=479232\alpha ^4-1217664\alpha ^3+1107600\alpha ^2-426864\alpha +60324>0$ are true for $\alpha \in (\alpha _2,1).$ Hence, the square trinomial $\varphi$ has two roots $t_{1,2}$ given by (3.11). Note that $t_2<0$ evidently. Moreover, the inequality $t_1<1$ is equivalent to

$$\begin{aligned} \begin{aligned}&\sqrt{479232\alpha ^4-1217664\alpha ^3+1107600\alpha ^2-426864\alpha +60324}\\&\quad <468\alpha ^3-258\alpha ^2-228\alpha +80, \end{aligned} \end{aligned}$$

which is equivalent to the inequality

$$\begin{aligned} 4(6\alpha +13)(234\alpha ^3-441\alpha ^2+276\alpha -61)(39\alpha ^2-54\alpha +17)>0 \end{aligned}$$

being true for $\alpha \in (\alpha _2,1).$

Thus, we conclude that for $\alpha \in (\alpha _2,1)$ the inequality (3.9) is true for $0<\zeta _1\le \zeta _1''$ where $\zeta _1'':=\sqrt{t_1}.$

Applying Lemma 2.2 for $0<\zeta _1\le \zeta _1''$, we get

$$\begin{aligned} \left| \Gamma _1\Gamma _3-\Gamma _2^2\right| \le \frac{1}{8}(1-\alpha )^2\zeta _1(1-\zeta _1^2)(-|A|+|B|+|C|)=:\mu (\zeta _1), \end{aligned}$$

where for $t\in \mathbb R,$

$$\begin{aligned} \mu (t):=-\frac{1}{144}(1-\alpha )^2\left( (39\alpha ^2-30\alpha -1)t^4-2(-17+12\alpha )t^2-16\right) . \end{aligned}$$

Since

$$\begin{aligned} \mu '(t)=-\frac{1}{36}(1-\alpha )^2 t\left( (39\alpha ^2-30\alpha -1)t^2+17-12\alpha )\right) <0 \end{aligned}$$

for $0<t\le \zeta _1'',$ we see that

$$\begin{aligned} \mu (\zeta _1)\le \mu (0)=\frac{1}{9}(1-\alpha )^2,\quad 0<\zeta _1\le \zeta _1''. \end{aligned}$$

CII.5 (a) Let $\alpha \in (\alpha _2,1).$ Applying Lemma 2.2 for $\zeta _1''<\zeta _1<1$, we get

$$\begin{aligned} |\Gamma _1\Gamma _3-\Gamma _2^2| \le \frac{1}{8}(1-\alpha )^2\zeta _1(1-\zeta _1^2)\left( |A|+|C| \right) \sqrt{1-\frac{B^2}{4AC}}=\psi (\zeta _1), \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} \psi (t):=&\frac{1}{144}(1-\alpha )^2\left( 3(13\alpha -5)(\alpha -1)t^4-14t^2+16\right) \\&\times \sqrt{\frac{3\left( -11(\alpha -1)^2t^2+128\alpha ^2-184\alpha +62\right) }{(39\alpha ^2-54\alpha +17)(t^2+8)}},\quad t\in [0,1]. \end{aligned} \end{aligned}$$

Since $-11(\alpha -1)^2t^2+128\alpha ^2-184\alpha +62>0$ and $(39\alpha ^2-54\alpha +17)(t^2+8)>0$ for $t\in [0,1],$ the function $\psi$ is well-defined. We have

$$\begin{aligned} \begin{aligned} \psi (\zeta _1'')=-\frac{(39\alpha ^2-54\alpha +17)(\alpha -1)^2}{18(234\alpha ^3-441\alpha ^2+276\alpha -61)^2}G_0(\alpha )\sqrt{-\frac{H_0(\alpha )}{K_0(\alpha )}} \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} G_0(\alpha )&:=-29952\alpha ^4+71892\alpha ^3-60480\alpha ^2+20907\alpha -2592+(78\alpha ^2-87\alpha +16)\\&\quad \times \sqrt{119808\alpha ^4-304416\alpha ^3+276900\alpha ^2-106716\alpha +15081},\\ \end{aligned} \end{aligned}$$

$$\begin{aligned} \begin{aligned} H_0(\alpha )&:=3(-29952\alpha ^5+96072\alpha ^4-119826\alpha ^3+72811\alpha ^2-21824\alpha +2671\\&\quad +(11\alpha ^2-22\alpha +11)\\&\quad \left. \times \sqrt{119808\alpha ^4-304416\alpha ^3+276900\alpha ^2-106716\alpha +15081}\right) \end{aligned} \end{aligned}$$

$$\begin{aligned} \begin{aligned} K_0(\alpha )&:=(39\alpha ^2-54\alpha +17)(1872\alpha ^3-3840\alpha ^2+2598\alpha -589\\&\quad \left. +\sqrt{119808\alpha ^4-304416\alpha ^3+276900\alpha ^2-106716\alpha +15081}\right) \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \psi (1)=\frac{1}{144}(\alpha -1)^2(39\alpha ^2-54\alpha +17). \end{aligned}$$

Differentiating $\psi$ lead to the equation

$$\begin{aligned} \begin{aligned} \psi '(t)=\frac{-(1-\alpha )^2tQ(t^2)}{24(39\alpha ^2-54\alpha +17)(t^2+8)^2\sqrt{\dfrac{-3(11(\alpha -1)^2t^2-128\alpha ^2+184\alpha -62)}{(t^2+8)(39\alpha ^2-54\alpha +17)}}}=0, \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} Q(s)&:=66s^3(13\alpha -5)(\alpha -1)^3+s^2(\alpha -1)(1092\alpha ^3-6816\alpha ^2+7259\alpha -1751)\\&\quad -2s(39936\alpha ^4-112704\alpha ^3+114668\alpha ^2-50068\alpha +8147)\\&\quad +16064\alpha ^2-23488\alpha +8144,\quad s\in [0,1]. \end{aligned} \end{aligned}$$

Now we describe the numbers of zeros of Q in the interval (0, 1) by combining Descartes’ and Laguerre’s rules. To apply Descartes’ rule, we check the numbers of sign changes of coefficients of the polynomial Q. We have:

• $u_0(\alpha ):=66(13\alpha -5)(\alpha -1)^3>0$ iff $\alpha \in \left( 0,5/13\right) ,$

• $u_1(\alpha ):=(\alpha -1)(1092\alpha ^3-6816\alpha ^2+7259\alpha -1751)>0$ iff $\alpha \in \left( 0,\alpha _4\right) \cup (\alpha _5,1),$ where $\alpha _4\approx 0.349478$ and $\alpha _5\approx 0.923387$,

• $u_2(\alpha ):=-2(39936\alpha ^4-112704\alpha ^3+114668\alpha ^2-50068\alpha +8147)>0$ iff $\alpha \in \left( \alpha _6,1\right) ,$ where $\alpha _6\approx 0.907318,$

• $u_3(\alpha ):=16064\alpha ^2-23488\alpha +8144 >0$ iff $\alpha \in \left( 0,\alpha _7\right) \cup (\alpha _8,1),$ where

  $$\begin{aligned} \alpha _7:=\frac{1}{502}(367-3\sqrt{770})\approx 0.565246 \end{aligned}$$

  and

  $$\begin{aligned} \alpha _8:=\frac{1}{502}(367+3\sqrt{770})\approx 0.896906. \end{aligned}$$

Thus, there is no change in signs in $\left( \alpha _7,\alpha _8\right)$, i.e., $N(Q,0)=0,$ one change of signs in $\left( 5/13,\alpha _7\right) \cup (\alpha _8,1),$ i.e., $N(Q,0)=1$ and two changes of signs in $\left( 0,5/13\right) ,$ i.e., $N(Q,0)=2.$ According to Descartes’ rule of signs, the polynomial Q has no positive real root in $\left( \alpha _7,\alpha _8\right)$, one positive real root in $\left( 5/13,\alpha _7\right) \cup (\alpha _8,1)$ and zero or two positive real roots in $\left( 0,5/13\right) .$ To apply Laguerres’ rule, it remains to compute the number N(Q, 1) of sign changes in the sequence of sums $\sum _{j=0}^ku_j(\alpha )$, where $k=0,\ldots 3.$ We have

• $u_0(\alpha )=66(13\alpha -5)(\alpha -1)^3>0$ iff $\alpha \in \left( 0,5/13\right) ,$

• $u_0(\alpha )+u_1(\alpha )=(\alpha - 1) (1950 \alpha ^3 - 8862 \alpha ^2 + 8777\alpha - 2081)>0$ iff $\alpha \in \left( 0,\alpha _9\right) \cup (\alpha _{10},1),$ where $\alpha _9\approx 0.353379$ and $\alpha _{10}\approx 0.924429,$

• $u_0(\alpha )+u_1(\alpha )+u_2(\alpha )=-(77922\alpha ^4 - 214596\alpha ^3 + 211697\alpha ^2 - 89278\alpha + 14213)>0$ iff $\alpha \in \left( 0,\alpha _{11}\right) ,$ where $\alpha _{11}\approx 0.90909,$

• $u_0(\alpha )+u_1(\alpha )+u_2(\alpha )+u_3(\alpha )= -77922\alpha ^4 + 214596\alpha ^3 - 195633\alpha ^2 + 65790\alpha - 6069=-3 (39\alpha ^2 - 54\alpha + 17) (666\alpha ^2 - 912\alpha + 119)>0$ iff $\alpha \in \left( \alpha _{12},\alpha _1\right) \cup (\alpha _2,1),$ where

  $$\begin{aligned} \alpha _{12}:=\frac{1}{111}(76-3\sqrt{(2383)/6})\approx 0.14606. \end{aligned}$$

Thus, there is no changes of signs in $\left( \alpha _1,\alpha _2\right)$, i.e., $N(Q,1)=0$, one change of sign in $\left( 0,\alpha _{12}\right) \cup \left( 5/13,\alpha _1\right) \cup (\alpha _2,1),$ i.e., $N(Q,1)=1,$ and two changes of sign in $\left( \alpha _{12},5/13\right) ,$ i.e., $N(Q,1)=2.$

According to Laguerre’s rule, the polynomial Q has one root in [0, 1] for $\alpha \in \left( 0,\alpha _{12}\right) \cup (\alpha _1,\alpha _{7})\cup (\alpha _8,\alpha _2)$ and no roots in [0, 1] for $\alpha \in \left( \alpha _{12},\alpha _1\right) \cup (\alpha _7,\alpha _8)\cup (\alpha _2,1).$

Therefore, for $\alpha \in \left( \alpha _2,1\right)$, the function $\psi$ is decreasing for $\zeta _1^0<t<1$ and hence

$$\begin{aligned} \psi (t)\le \psi (\zeta _1^0),\quad \zeta _1^0<t<1. \end{aligned}$$

(b) Let $\alpha \in (0,\alpha _1).$ Applying Lemma 2.2 for $\zeta _1'<\zeta _1<1$, we get the same function like as in CII.5(a) and therefore we repeat the considerations. We have

$$\begin{aligned} \begin{aligned} \psi (\zeta _1')=-\frac{39\alpha ^2-54\alpha +17}{54(\alpha -1)(78\alpha -43)^2}G_1(\alpha )\sqrt{-\frac{H_1(\alpha )}{K_1(\alpha )}} \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} G_1(\alpha )&:=-9984\alpha ^3+22404\alpha ^2-16440\alpha +3939\\&\quad +(26\alpha -17) \sqrt{119808\alpha ^4-371808\alpha ^3+426372\alpha ^2-213852\alpha +39561} \end{aligned} \end{aligned}$$

$$\begin{aligned} \begin{aligned} H_1(\alpha )&:=3(\alpha -1)^2(29952\alpha ^3-63000\alpha ^2+43458\alpha -9879\\&\quad +11\sqrt{119808\alpha ^4-371808\alpha ^3+426372\alpha ^2-213852\alpha +39561}) \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} K_1(\alpha )&:=(39\alpha ^2-54\alpha +17)(-1872\alpha ^3+4464\alpha ^2-3462\alpha +861\\&\quad +\sqrt{119808\alpha ^4-371808\alpha ^3+426372\alpha ^2-213852\alpha +39561}). \end{aligned} \end{aligned}$$

For $\alpha \in \left( 0,\alpha _{12}\right)$, the function $\psi$ has a unique critical point in [0, 1],  where using jointly Descartes’ and Laguerre’s rules we state that $\psi$ attains its minimum value. Thus,

$$\begin{aligned} \psi (t)\le \max \{\psi (\zeta _1'),\psi (1)\},\quad \zeta _1'<t<1. \end{aligned}$$

For $\alpha \in \left( \alpha _{12},\alpha _1\right)$, the function $\psi$ is decreasing for $\zeta _1'<t<1$ and hence

$$\begin{aligned} \psi (t)\le \psi (\zeta _1'),\quad \zeta _1'<t<1. \end{aligned}$$

CIII. Now we summarize results of Parts CI-CII.

(i) For $\alpha \in \left( 0,\alpha _{12}\right] ,$ we compare $\psi (1)$ and $\sigma (t_5).$ Note that the inequality

$$\begin{aligned} \psi (1)=\frac{1}{144}(\alpha -1)^2(39\alpha ^2-54\alpha +17)\ge \frac{1}{1872}(256\alpha ^2-440\alpha +211)=\sigma (t_5), \end{aligned}$$

is equivalent to

$$\begin{aligned} \frac{1}{1872}(507\alpha ^4 - 1716\alpha ^3 + 1876\alpha ^2 - 704\alpha + 10)\ge 0, \end{aligned}$$

which is true for $\alpha \in (0,\alpha _0],$ where $\alpha _0\approx 0.014779.$ Thus, $\psi (1)\ge \sigma (t_5)$ for $\alpha \in (0,\alpha _0),$ and $\psi (1)\le \sigma (t_5)$ for $\alpha \in [\alpha _0,\alpha _{12}].$

(ii) For $\alpha \in \left( \alpha _{12},1/4\right] ,$ we have

$$\begin{aligned} \left| \Gamma _1\Gamma _3-\Gamma _2^2\right| \le \max \{\sigma (t_5),\psi (\zeta _1')\}=\sigma (t_5). \end{aligned}$$

(iii) For $\alpha \in \left( 1/4,\alpha _1\right) ,$ we have

$$\begin{aligned} \left| \Gamma _1\Gamma _3-\Gamma _2^2\right| \le \max \{\sigma (0),\psi (\zeta _1')\}=\sigma (0)=\frac{1}{9}(1-\alpha )^2. \end{aligned}$$

(iv) For $\alpha \in \left[ \alpha _1,\alpha _2\right] ,$ we have

$$\begin{aligned} \left| \Gamma _1\Gamma _3-\Gamma _2^2\right| \le \rho (0)=\frac{1}{9}(1-\alpha )^2 \end{aligned}$$

(v) For $\alpha \in \left( \alpha _2,1\right) ,$ we have

$$\begin{aligned} \left| \Gamma _1\Gamma _3-\Gamma _2^2\right| \le \max \{\mu (0),\psi (\zeta _1'')\}=\mu (0)=\frac{1}{9}(1-\alpha )^2. \end{aligned}$$

It remains to show sharpness of all inequalities.

Equality in the first inequality in (3.1) holds for the function $f\in \mathcal P'(\alpha )$ given by (3.2) with

$$\begin{aligned} p(z):=\frac{1+z}{1-z},\quad z\in \mathbb D, \end{aligned}$$

having by (3.3) coefficients

$$\begin{aligned} a_2(f)=1-\alpha ,\quad a_3(f)=\frac{2}{3}(1-\alpha ),\quad a_4(f)=\frac{1}{2}(1-\alpha ). \end{aligned}$$

Equality in the second inequality in (3.1) holds for the function $f\in \mathcal P'(\alpha )$ given by (3.2), where $p\in \mathcal P$ is defined by (2.5) with $\zeta _1=t_5=:\tau$ and $\zeta _2=1,$ i.e.,

$$\begin{aligned} p(z) = \frac{1+2\tau z + z^2}{1-z^2}, \quad z\in {\mathbb {D}}. \end{aligned}$$

Then by (3.3) the function f has the following coefficients:

$$\begin{aligned} a_2(f)=(1-\alpha )\tau ,\quad a_3(f)=\frac{2}{3}(1-\alpha ),\quad a_4(f)=\frac{1}{2}(1-\alpha )\tau . \end{aligned}$$

Equality in the third inequality in (3.1) holds for the function $f\in \mathcal P'(\alpha )$ given by (3.2) with

$$\begin{aligned} p(z):=\frac{1+z^2}{1-z^2},\quad z\in \mathbb D, \end{aligned}$$

having by (3.3) coefficients

$$\begin{aligned} a_2(f)=a_4(f)=0,\quad a_3(f)=\frac{2}{3}(1-\alpha ). \end{aligned}$$

This ends the proof of the theorem. $\square$

For $\alpha =0$, we have

Corollary 3.1

If $f\in {{\mathcal {P}}'},$ then

$$\begin{aligned} \left| \Gamma _1\Gamma _3-\Gamma _2^2\right| \le \frac{17}{144}. \end{aligned}$$

The inequality is sharp.