A Resolution of an Open Problem on the Riemann Zeta Function via Tail Bounds and Zeta-Ratio Monotonicity

1. Introduction

The Riemann zeta function is defined, for $s>1$, by

$$\zeta \left(s\right)={\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n^s}}.$$

It is one of the central functions of analytic number theory and has important connections with special functions, probability theory, mathematical physics, and the theory of inequalities.

A considerable body of work has been devoted to inequalities, convexity properties, monotonicity, and ratio estimates associated with the Riemann zeta function and related Dirichlet functions. Cerone and Dragomir [2] investigated several inequalities involving the Riemann zeta function and related convexity properties. Chen, Guo, and Wang [3] studied logarithmic convexity and log-behaviour of sequences arising from zeta functions. Of particular relevance are inequalities and monotonicity properties involving ratios of zeta values. Yang and Tian [4] obtained sharp bounds for ratios of two Riemann zeta functions. Guo and Qi [6] studied increasing properties and logarithmic convexity of functions involving the Riemann zeta function, including functions of the form

$$\left({\displaystyle \genfrac{}{}{0pt}{}{x+\alpha +\ell }{\alpha }}\right){\displaystyle \frac{\zeta (x+\alpha )}{\zeta \left(x\right)}}.$$

They established monotonicity using integral representations and monotonicity rules for ratios of parameter-dependent integrals. Lim and Qi [7] and Qi and Lim [8] studied analogous properties for Dirichlet eta and lambda functions.

Against this background, Nantomah and Ravi [1] proved that, for every positive integer n,

$$(n+2)\zeta (n+1)\zeta (n+3)-(n+1){\zeta }^2(n+2)-\zeta (n+1)\zeta (n+2)>0,$$

and proposed the corresponding extension

$$(z+2)\zeta (z+1)\zeta (z+3)-(z+1){\zeta }^2(z+2)-\zeta (z+1)\zeta (z+2)>0,z>0,$$

as an open problem.

Define

$$F\left(z\right)=(z+2)\zeta (z+1)\zeta (z+3)-(z+1){\zeta }^2(z+2)-\zeta (z+1)\zeta (z+2).$$

The purpose of the present paper is to investigate the positivity of F on $(0,\infty )$.

The principal contributions of this paper are threefold. First, we derive the explicit lower bound

$$F\left(z\right)\ge \left(z-\tau (z+1)\right)\left(\tau (z+2)-\tau (z+3)\right)$$

by combining a direct series factorization with the log-convexity of the tail function $\tau =\zeta -1$. This yields an independent proof of positivity on the interval

$$z>z_0,$$

where $z_0$ is the unique positive solution of

$$\zeta (z_0+1)=z_0+1.$$

Second, we reformulate the full problem exactly as a unit-step inequality for the function

$$H\left(x\right)=x\left(1-{\displaystyle \frac{\zeta (x+1)}{\zeta \left(x\right)}}\right).$$

Third, we connect this formulation with contemporary monotonicity results for ratios of shifted zeta values. In particular, an application of a two-variable monotonicity theorem of Yang and Tian leads to a stronger consecutive-ratio bound that yields the desired positivity throughout the full range $z>0$.

Our first approach is based on the tail function

$$\tau \left(s\right)=\zeta \left(s\right)-1={\displaystyle \sum _{n=2}^{\infty }}{\displaystyle \frac{1}{n^s}}.$$

Writing

$$\alpha =z+1$$

and introducing

$$A=\tau \left(\alpha \right),B=\tau (\alpha +1),C=\tau (\alpha +2),$$

we decompose $F\left(z\right)$ into a linear part L and a quadratic part T. A direct series factorization gives a lower estimate for L, whereas the log-convexity of $\tau$ provides the Turán-type estimate

$$B^2\le AC,$$

which is used to control T.

Combining these estimates yields the fundamental inequality

$$F\left(z\right)\ge \left(z-\tau (z+1)\right)\left(\tau (z+2)-\tau (z+3)\right),z>0.$$

Since $\tau$ is strictly decreasing, the second factor is strictly positive. Consequently,

$$\zeta (z+1)<z+1$$

is a sufficient condition for $F\left(z\right)>0$.

We then analyze the equation

$$\zeta (z+1)=z+1$$

and prove that it has a unique positive solution $z_0\in (0,1)$. Consequently, the desired inequality holds for every

$$z>z_0.$$

A numerical solution of the defining equation gives

$$z_0\approx 0.8337726517.$$

The numerical approximation is included only to indicate the location of the threshold and is not required in the proof.

In particular, the present argument strengthens the initial range

$$z\ge 1$$

to the larger interval

$$z>z_0.$$

Thus the present lower-bound method establishes positivity below $z=1$. The interval not covered by this method is

$$0<z\le z_0.$$

No conclusion concerning the sign of $F\left(z\right)$ on this interval follows merely from the failure of the present lower bound to be positive there.

Finally, motivated in part by contemporary work on ratios of zeta values, we introduce

$$H\left(x\right)=x\left(1-{\displaystyle \frac{\zeta (x+1)}{\zeta \left(x\right)}}\right),x>1.$$

We establish the exact identity

$$F\left(z\right)=\zeta (z+1)\zeta (z+2)\left(H(z+1)-H(z+2)\right).$$

It follows that the original inequality for all $z>0$ is equivalent to the unit-step decrease condition

$$H\left(x\right)>H(x+1),x>1.$$

Accordingly, the outstanding part of the problem is reduced to establishing

$$H\left(x\right)>H(x+1),1<x\le z_0+1.$$

Strict monotonicity of H on $(1,\infty )$ would provide a stronger sufficient condition, but is not necessary for resolving the original unit-step inequality.

2. Preliminaries

For $s>1$, define

$$\tau \left(s\right)=\zeta \left(s\right)-1={\displaystyle \sum _{n=2}^{\infty }}{\displaystyle \frac{1}{n^s}}.$$

Clearly,

$$\tau \left(s\right)>0,s>1.$$

Lemma 1

(Strict monotonicity of the tail function). The function τ is strictly decreasing on $(1,\infty )$.

Proof. 

Let $s_2>s_1>1$. For every integer $n\ge 2$,

$$n^{-s_2}<n^{-s_1}.$$

Since the corresponding series converge absolutely, summing over $n\ge 2$ gives

$$\tau \left(s_2\right)<\tau \left(s_1\right).$$

Therefore $\tau$ is strictly decreasing on $(1,\infty )$. □

Lemma 2

(Log-convexity of the tail function). The function

$$\tau \left(s\right)={\displaystyle \sum _{n=2}^{\infty }}{n}^{-s}$$

is log-convex on $(1,\infty )$. In particular, for every $\alpha >1$,

$$\tau {(\alpha +1)}^2\le \tau \left(\alpha \right)\tau (\alpha +2).$$

(2.1)

Proof. 

Let $s,t>1$ and $0<\lambda <1$. Then

$$\begin{array}{cc}\hfill \tau (\lambda s+(1-\lambda \left)t\right)& ={\displaystyle \sum _{n=2}^{\infty }}{n}^{-\lambda s-(1-\lambda )t}\hfill \\ & ={\displaystyle \sum _{n=2}^{\infty }}{\left(n^{-s}\right)}^{\lambda }{\left(n^{-t}\right)}^{1-\lambda }.\hfill \end{array}$$

Applying Hölder’s inequality with conjugate exponents

$$p={\displaystyle \frac{1}{\lambda }},q={\displaystyle \frac{1}{1-\lambda }},$$

we obtain

$$\tau (\lambda s+(1-\lambda )t)\le {\left({\displaystyle \sum _{n=2}^{\infty }}{n}^{-s}\right)}^{\lambda }{\left({\displaystyle \sum _{n=2}^{\infty }}{n}^{-t}\right)}^{1-\lambda }.$$

Therefore,

$$\tau (\lambda s+(1-\lambda )t)\le \tau {\left(s\right)}^{\lambda }\tau {\left(t\right)}^{1-\lambda },$$

which proves that $\tau$ is log-convex on $(1,\infty )$.

Taking

$$s=\alpha ,t=\alpha +2,\lambda ={\displaystyle \frac{1}{2}},$$

we obtain

$$\tau (\alpha +1)\le \sqrt{\tau \left(\alpha \right)\tau (\alpha +2)}.$$

Since all terms are positive, squaring both sides yields (2.1). □

Remark 1.

The preceding proof establishes the log-convexity of $\tau \left(s\right)=\zeta \left(s\right)-1$ directly from its Dirichlet series representation. Thus no inference from the log-convexity of ζ itself is required. This distinction is important because log-convexity is not, in general, preserved under subtraction of a positive constant.

Remark 2

(A Turán-type interpretation). The inequality (2.1) is the discrete midpoint inequality associated with the log-convexity of τ. It may therefore be regarded as a Turán-type inequality with the orientation characteristic of log-convex functions.

Lemma 3

(Guo–Qi weighted shifted-ratio monotonicity [6]). For $\alpha >0$ and $\ell \in {\mathbb{N}}_0$, the function

$$x⟼\left({\displaystyle \genfrac{}{}{0pt}{}{x+\alpha +\ell }{\alpha }}\right){\displaystyle \frac{\zeta (x+\alpha )}{\zeta \left(x\right)}}$$

is strictly increasing on $(1,\infty )$.

Proof. 

This is a result of Guo and Qi [6]. The proof uses an integral representation of the ratio and a monotonicity rule for ratios of parameter-dependent integrals. We refer to the original source for details. □

Corollary 1.

For $\alpha =1$, $\ell =0$, the function

$$x⟼(x+1){\displaystyle \frac{\zeta (x+1)}{\zeta \left(x\right)}}$$

is strictly increasing on $(1,\infty )$. Consequently, for $x>1$,

$$(x+2)r(x+1)>(x+1)r\left(x\right),$$

where $r\left(x\right)={\displaystyle \frac{\zeta (x+1)}{\zeta \left(x\right)}}$. Hence

$$r(x+1)>{\displaystyle \frac{x+1}{x+2}}r\left(x\right).$$

(2.2)

Theorem 1

(Yang–Tian two-variable monotonicity [5]). Let

$$\mathsf{\Phi }(x,y)={\displaystyle \frac{2^{-x}\zeta \left(x\right)-2^{-y}\zeta \left(y\right)}{\zeta \left(x\right)-\zeta \left(y\right)}},x,y>1,x\ne y.$$

Assume that Φ is defined on the diagonal by its continuous extension. Then Φ is strictly increasing with respect to each variable on ${(1,\infty )}^2$.

Remark 3.

The exact normalization of Φ, the treatment of the diagonal $x=y$, and the precise monotonicity hypotheses in Theorem 1 are stated as in the original work of Yang and Tian [5]. The theorem is used in the form presented above.

3. Results and Discussion

3.1. Algebraic Decomposition and Preliminary Estimates

Let

$$\alpha =z+1,z>0,$$

so that $\alpha >1$. Define

$$A=\tau \left(\alpha \right),B=\tau (\alpha +1),C=\tau (\alpha +2).$$

Lemma 4

(Algebraic decomposition). For every $z>0$,

$$F\left(z\right)=L+T,$$

where

$$L=\alpha A+(\alpha +1)C-(2\alpha +1)B$$

and

$$T=(\alpha +1)AC-\alpha B^2-AB.$$

Proof. 

Since

$$z+1=\alpha ,z+2=\alpha +1,z+3=\alpha +2,$$

we have

$$F\left(z\right)=(\alpha +1)\zeta \left(\alpha \right)\zeta (\alpha +2)-\alpha {\zeta }^2(\alpha +1)-\zeta \left(\alpha \right)\zeta (\alpha +1).$$

Now

$$\zeta \left(\alpha \right)=1+A,\zeta (\alpha +1)=1+B,\zeta (\alpha +2)=1+C.$$

Therefore

$$\begin{array}{cc}\hfill F\left(z\right)& =(\alpha +1)(1+A)(1+C)-\alpha {(1+B)}^2-(1+A)(1+B).\hfill \end{array}$$

Expanding and collecting the linear and quadratic terms gives

$$F\left(z\right)=\alpha A+(\alpha +1)C-(2\alpha +1)B+(\alpha +1)AC-\alpha B^2-AB.$$

Hence

$$F\left(z\right)=L+T.$$

□

Lemma 5

(Lower bound for the linear part). For every $\alpha >1$,

$$L\ge (\alpha -1)(B-C).$$

(3.1)

Proof. 

Using the series representations of A, B, and C, we obtain

$$\begin{array}{cc}\hfill L& ={\displaystyle \sum _{n=2}^{\infty }}\left({\displaystyle \frac{\alpha }{n^{\alpha }}}-{\displaystyle \frac{2\alpha +1}{n^{\alpha +1}}}+{\displaystyle \frac{\alpha +1}{n^{\alpha +2}}}\right)\hfill \\ & ={\displaystyle \sum _{n=2}^{\infty }}{\displaystyle \frac{\alpha n^2-(2\alpha +1)n+\alpha +1}{n^{\alpha +2}}}.\hfill \end{array}$$

The numerator factors as

$$\alpha n^2-(2\alpha +1)n+\alpha +1=(n-1)\left(\alpha (n-1)-1\right).$$

Thus

$$L={\displaystyle \sum _{n=2}^{\infty }}{\displaystyle \frac{(n-1)\left(\alpha (n-1)-1\right)}{n^{\alpha +2}}}.$$

Since $n\ge 2$,

$$\alpha (n-1)-1\ge \alpha -1.$$

Therefore

$$\begin{array}{cc}\hfill L& \ge (\alpha -1){\displaystyle \sum _{n=2}^{\infty }}{\displaystyle \frac{n-1}{n^{\alpha +2}}}\hfill \\ & =(\alpha -1){\displaystyle \sum _{n=2}^{\infty }}\left({\displaystyle \frac{1}{n^{\alpha +1}}}-{\displaystyle \frac{1}{n^{\alpha +2}}}\right)\hfill \\ & =(\alpha -1)(B-C).\hfill \end{array}$$

□

Lemma 6

(Lower bound for the quadratic part). For every $\alpha >1$,

$$T\ge -A(B-C).$$

(3.2)

Proof. 

By Lemma 2,

$$B^2=\tau {(\alpha +1)}^2\le \tau \left(\alpha \right)\tau (\alpha +2)=AC.$$

Since $\alpha >0$,

$$-\alpha B^2\ge -\alpha AC.$$

Consequently,

$$\begin{array}{cc}\hfill T& =(\alpha +1)AC-\alpha B^2-AB\hfill \\ & \ge (\alpha +1)AC-\alpha AC-AB\hfill \\ & =AC-AB\hfill \\ & =-A(B-C).\hfill \end{array}$$

□

3.2. Fundamental Lower Bound

Theorem 2

(Fundamental lower bound). For every $z>0$,

$$F\left(z\right)\ge \left(z-\tau (z+1)\right)\left(\tau (z+2)-\tau (z+3)\right).$$

(3.3)

Proof. 

By Lemma 4,

$$F\left(z\right)=L+T.$$

Lemmas 5 and 6 give

$$L\ge (\alpha -1)(B-C)$$

and

$$T\ge -A(B-C).$$

Hence

$$\begin{array}{cc}\hfill F\left(z\right)& \ge (\alpha -1)(B-C)-A(B-C)\hfill \\ & =(\alpha -1-A)(B-C).\hfill \end{array}$$

Since

$$\alpha =z+1,A=\tau (z+1),$$

and

$$B-C=\tau (z+2)-\tau (z+3),$$

we obtain (3.3). □

Corollary 2

(A sufficient condition). If $z>0$ satisfies

$$\zeta (z+1)<z+1,$$

then

$$F\left(z\right)>0.$$

Proof. 

By Lemma 1,

$$\tau (z+2)-\tau (z+3)>0.$$

Moreover,

$$\zeta (z+1)<z+1$$

is equivalent to

$$\tau (z+1)<z.$$

Thus

$$z-\tau (z+1)>0.$$

The conclusion follows from Theorem 2. □

3.3. Threshold Analysis

We now determine precisely the range obtained from the sufficient condition in Corollary 2.

Define

$$g\left(z\right)=z+1-\zeta (z+1),z>0.$$

Lemma 7.

The function g is strictly increasing on $(0,\infty )$.

Proof. 

For $s>1$,

$${\zeta }^{\prime }\left(s\right)=-{\displaystyle \sum _{n=2}^{\infty }}{\displaystyle \frac{logn}{n^s}}<0.$$

Therefore

$$g^{\prime }\left(z\right)=1-{\zeta }^{\prime }(z+1)>1>0.$$

Hence g is strictly increasing on $(0,\infty )$. □

Theorem 3

(Existence and uniqueness of the threshold). There exists a unique $z_0\in (0,1)$ such that

$$\zeta (z_0+1)=z_0+1.$$

(3.4)

Moreover,

$$\zeta (z+1)<z+1$$

for every $z>z_0$.

Proof. 

Since $\zeta \left(s\right)\to +\infty$ as $s\to 1^{+}$,

$$\underset{z\to 0^{+}}{lim}\left(z+1-\zeta (z+1)\right)=-\infty .$$

On the other hand,

$$g\left(1\right)=2-\zeta \left(2\right)=2-{\displaystyle \frac{{\pi }^2}{6}}>0.$$

By continuity, there exists at least one $z_0\in (0,1)$ satisfying

$$g\left(z_0\right)=0.$$

By Lemma 7, g is strictly increasing, so this zero is unique.

Furthermore, if $z>z_0$, then

$$g\left(z\right)>g\left(z_0\right)=0.$$

Therefore

$$z+1-\zeta (z+1)>0,$$

or equivalently,

$$\zeta (z+1)<z+1.$$

□

Theorem 4

(Improved partial resolution). Let $z_0$ denote the unique positive solution of (3.4). Then

$$(z+2)\zeta (z+1)\zeta (z+3)-(z+1){\zeta }^2(z+2)-\zeta (z+1)\zeta (z+2)>0$$

for every

$$z>z_0.$$

Numerically,

$$z_0\approx 0.8337726517.$$

Proof. 

Let $z>z_0$. By Theorem 3,

$$\zeta (z+1)<z+1.$$

Therefore, by Corollary 2,

$$F\left(z\right)>0.$$

□

Corollary 3.

Let $z_0$ denote the unique positive solution of (3.4). Then the inequality

$$(z+2)\zeta (z+1)\zeta (z+3)-(z+1){\zeta }^2(z+2)-\zeta (z+1)\zeta (z+2)>0$$

holds for every

$$z>z_0.$$

Consequently, the interval not covered by the present lower-bound method is

$$0<z\le z_0.$$

Remark 4

(Numerical value of the threshold). The threshold $z_0$ is defined analytically as the unique solution of (3.4). Numerically solving this equation gives

$$z_0\approx 0.8337726517.$$

The proof of the positivity result on $(z_0,\infty )$ depends only on the existence and uniqueness of $z_0$ and not on its numerical approximation.

3.4. Equivalent Unit-Step Formulation

Define

$$H\left(x\right)=x\left(1-{\displaystyle \frac{\zeta (x+1)}{\zeta \left(x\right)}}\right),x>1.$$

Theorem 5

(Equivalent unit-step formulation). For every $z>0$,

$$F\left(z\right)=\zeta (z+1)\zeta (z+2)\left(H(z+1)-H(z+2)\right).$$

(3.5)

Consequently,

$$F\left(z\right)>0⟺H(z+1)>H(z+2).$$

Equivalently, the original inequality holds for every $z>0$ if and only if

$$H\left(x\right)>H(x+1),x>1.$$

(3.6)

Proof. 

Set

$$\alpha =z+1.$$

Then

$$F\left(z\right)=(\alpha +1)\zeta \left(\alpha \right)\zeta (\alpha +2)-\alpha {\zeta }^2(\alpha +1)-\zeta \left(\alpha \right)\zeta (\alpha +1).$$

Since

$$\zeta \left(\alpha \right)\zeta (\alpha +1)>0,$$

we may write

$$\begin{array}{cc}\hfill F\left(z\right)& =\zeta \left(\alpha \right)\zeta (\alpha +1)\left[(\alpha +1){\displaystyle \frac{\zeta (\alpha +2)}{\zeta (\alpha +1)}}-\alpha {\displaystyle \frac{\zeta (\alpha +1)}{\zeta \left(\alpha \right)}}-1\right].\hfill \end{array}$$

On the other hand,

$$H\left(\alpha \right)=\alpha -\alpha {\displaystyle \frac{\zeta (\alpha +1)}{\zeta \left(\alpha \right)}}$$

and

$$H(\alpha +1)=\alpha +1-(\alpha +1){\displaystyle \frac{\zeta (\alpha +2)}{\zeta (\alpha +1)}}.$$

Therefore,

$$\begin{array}{cc}\hfill H\left(\alpha \right)-H(\alpha +1)& =(\alpha +1){\displaystyle \frac{\zeta (\alpha +2)}{\zeta (\alpha +1)}}-\alpha {\displaystyle \frac{\zeta (\alpha +1)}{\zeta \left(\alpha \right)}}-1.\hfill \end{array}$$

Hence

$$F\left(z\right)=\zeta \left(\alpha \right)\zeta (\alpha +1)\left(H\left(\alpha \right)-H(\alpha +1)\right).$$

Substituting $\alpha =z+1$ yields (3.5).

Since

$$\zeta (z+1)\zeta (z+2)>0$$

for every $z>0$, we conclude that

$$F\left(z\right)>0⟺H(z+1)>H(z+2).$$

Finally, setting

$$x=z+1,$$

we have $x>1$, and hence the original inequality holds for every $z>0$ if and only if (3.6) holds. □

Corollary 4

(A sufficient monotonicity criterion). If

$$H\left(x\right)=x\left(1-{\displaystyle \frac{\zeta (x+1)}{\zeta \left(x\right)}}\right)$$

is strictly decreasing on $(1,\infty )$, then

$$F\left(z\right)>0$$

for every $z>0$.

Proof. 

Let $z>0$. Then

$$z+1<z+2.$$

If H is strictly decreasing on $(1,\infty )$, it follows that

$$H(z+1)>H(z+2).$$

Therefore, by Theorem 5,

$$F\left(z\right)>0.$$

□

Remark 5.

The exact condition equivalent to the original zeta-function inequality is the unit-step decrease condition (3.6). Strict monotonicity of H on $(1,\infty )$ is a stronger sufficient condition, but it is not necessary for the unit-step inequality. Thus, a complete solution of the original problem requires only the proof of (3.6).

3.5. Monotonicity Results for Shifted Zeta Ratios

Define

$$r\left(x\right)={\displaystyle \frac{\zeta (x+1)}{\zeta \left(x\right)}},x>1.$$

Since the Riemann zeta function is strictly decreasing on $(1,\infty )$,

$$0<r\left(x\right)<1.$$

The exact H-function formulation gives

$$F\left(z\right)>0$$

if and only if, with $x=z+1$,

$$(x+1)r(x+1)-xr\left(x\right)>1.$$

Equivalently,

$$r(x+1)>{\displaystyle \frac{1+xr\left(x\right)}{x+1}}.$$

(3.7)

We now compare this required estimate with known monotonicity results for shifted zeta ratios.

3.5.1. The Guo–Qi Weighted-Ratio Result

By Lemma 3 and its corollary, we have

$$r(x+1)>{\displaystyle \frac{x+1}{x+2}}r\left(x\right).$$

(3.8)

Although (3.8) provides nontrivial information about consecutive zeta ratios, it does not by itself imply (3.7). Thus the Guo–Qi theorem alone does not resolve the present problem.

3.5.2. A Two-Variable Zeta-Ratio Approach

By Theorem 1, the function

$$\mathsf{\Phi }(x,y)={\displaystyle \frac{2^{-x}\zeta \left(x\right)-2^{-y}\zeta \left(y\right)}{\zeta \left(x\right)-\zeta \left(y\right)}}$$

is strictly increasing in each variable. Define

$$Q\left(x\right)=\mathsf{\Phi }(x,x+1),x>1.$$

Then Q is strictly increasing on $(1,\infty )$, because for $x_2>x_1>1$,

$$Q\left(x_2\right)=\mathsf{\Phi }(x_2,x_2+1)>\mathsf{\Phi }(x_1,x_2+1)>\mathsf{\Phi }(x_1,x_1+1)=Q\left(x_1\right).$$

Hence

$$Q(x+1)>Q\left(x\right),x>1.$$

(3.9)

We now derive a stronger ratio bound using this fact.

Theorem 6

(Consecutive zeta-ratio bound). For every $x>1$,

$$r(x+1)>{\displaystyle \frac{2}{3-r\left(x\right)}}.$$

(3.10)

Proof. 

Let

$$a=r\left(x\right)={\displaystyle \frac{\zeta (x+1)}{\zeta \left(x\right)}}$$

and

$$b=r(x+1)={\displaystyle \frac{\zeta (x+2)}{\zeta (x+1)}}.$$

Since $\zeta$ is positive and strictly decreasing on $(1,\infty )$,

$$0<a<1,0<b<1.$$

From the definition of Q,

$$\begin{array}{cc}\hfill Q\left(x\right)& ={\displaystyle \frac{2^{-x}\zeta \left(x\right)-2^{-(x+1)}\zeta (x+1)}{\zeta \left(x\right)-\zeta (x+1)}}\hfill \\ & =2^{-x}{\displaystyle \frac{1-\frac{1}{2}a}{1-a}}\hfill \\ & =2^{-x}{\displaystyle \frac{2-a}{2(1-a)}}.\hfill \end{array}$$

Similarly,

$$Q(x+1)=2^{-(x+1)}{\displaystyle \frac{2-b}{2(1-b)}}.$$

By (3.9), $Q(x+1)>Q\left(x\right)$. Hence

$$2^{-(x+1)}{\displaystyle \frac{2-b}{2(1-b)}}>2^{-x}{\displaystyle \frac{2-a}{2(1-a)}}.$$

Multiplying by $2^{x+2}>0$ gives

$${\displaystyle \frac{2-b}{1-b}}>2{\displaystyle \frac{2-a}{1-a}}.$$

Since $0<a,b<1$, all denominators are positive, so cross-multiplication yields

$$(2-b)(1-a)>2(2-a)(1-b).$$

Expanding,

$$2-2a-b+ab>4-2a-4b+2ab.$$

Hence

$$3b-ab>2,$$

or

$$b(3-a)>2.$$

Since $3-a>0$,

$$b>{\displaystyle \frac{2}{3-a}}.$$

Substituting the definitions of a and b proves (3.10). □

3.6. Comparison with the Required Ratio Bound

Lemma 8.

Let $x>1$ and $0<a<1$. Then

$${\displaystyle \frac{2}{3-a}}>{\displaystyle \frac{1+xa}{x+1}}.$$

(3.11)

Proof. 

Since all denominators are positive, the desired inequality is equivalent to

$$2(x+1)>(3-a)(1+xa).$$

The difference between the two sides factors as

$$\begin{array}{cc}\hfill 2(x+1)-(3-a)(1+xa)& =2x+2-3-3xa+a+x{a}^2\hfill \\ & =(a-1)(ax-2x+1).\hfill \end{array}$$

Since

$$0<a<1,$$

we have

$$a-1<0.$$

Moreover,

$$ax-2x+1<x-2x+1=1-x<0,$$

because $a<1$ and $x>1$. Therefore,

$$(a-1)(ax-2x+1)>0.$$

It follows that

$$2(x+1)>(3-a)(1+xa),$$

and hence (3.11). □

3.7. Complete Positivity Theorem

Theorem 7

(Main theorem). For every $z>0$,

$$(z+2)\zeta (z+1)\zeta (z+3)-(z+1){\zeta }^2(z+2)-\zeta (z+1)\zeta (z+2)>0.$$

Proof. 

Let

$$x=z+1>1$$

and define

$$a=r\left(x\right)={\displaystyle \frac{\zeta (x+1)}{\zeta \left(x\right)}},b=r(x+1)={\displaystyle \frac{\zeta (x+2)}{\zeta (x+1)}}.$$

By Theorem 6,

$$b>{\displaystyle \frac{2}{3-a}}.$$

By Lemma 8,

$${\displaystyle \frac{2}{3-a}}>{\displaystyle \frac{1+xa}{x+1}}.$$

Therefore,

$$b>{\displaystyle \frac{1+xa}{x+1}}.$$

Multiplying by $x+1>0$ gives

$$(x+1)b>1+xa,$$

and hence

$$(x+1)b-xa-1>0.$$

Substituting the definitions of a and b gives

$$(x+1){\displaystyle \frac{\zeta (x+2)}{\zeta (x+1)}}-x{\displaystyle \frac{\zeta (x+1)}{\zeta \left(x\right)}}-1>0.$$

Multiplying by

$$\zeta \left(x\right)\zeta (x+1)>0$$

yields

$$(x+1)\zeta \left(x\right)\zeta (x+2)-x{\zeta }^2(x+1)-\zeta \left(x\right)\zeta (x+1)>0.$$

Finally, substituting

$$x=z+1$$

gives

$$(z+2)\zeta (z+1)\zeta (z+3)-(z+1){\zeta }^2(z+2)-\zeta (z+1)\zeta (z+2)>0.$$

This proves the result. □

Corollary 5.

For every $x>1$,

$$H\left(x\right)>H(x+1),$$

where

$$H\left(x\right)=x\left(1-{\displaystyle \frac{\zeta (x+1)}{\zeta \left(x\right)}}\right).$$

Proof. 

Since

$$H\left(x\right)=x(1-r(x\left)\right),$$

we have

$$\begin{array}{cc}\hfill H\left(x\right)-H(x+1)& =x(1-r(x\left)\right)-(x+1)(1-r(x+1\left)\right)\hfill \\ & =(x+1)r(x+1)-xr\left(x\right)-1.\hfill \end{array}$$

The proof of Theorem 7 shows that

$$(x+1)r(x+1)-xr\left(x\right)-1>0.$$

Therefore,

$$H\left(x\right)>H(x+1).$$

□

Remark 6.

The conclusion

$$H\left(x\right)>H(x+1)$$

is exactly the unit-step inequality required for the original problem. The argument does not require the stronger assertion that H be strictly decreasing for every pair $x_1<x_2$.

3.8. Discussion of the Two Approaches

The analysis developed in this paper reveals two complementary approaches to the original zeta-function inequality.

The first is the tail-function method. By introducing

$$\tau \left(s\right)=\zeta \left(s\right)-1,$$

we obtained the explicit lower bound (3.3). This result is independent of the ratio-monotonicity argument and provides a quantitative estimate for $F\left(z\right)$. It also yields the unconditional positivity result

$$F\left(z\right)>0,z>z_0,$$

where $z_0$ is the unique positive solution of (3.4).

The second approach is based on ratios of consecutive zeta values. Writing

$$r\left(x\right)={\displaystyle \frac{\zeta (x+1)}{\zeta \left(x\right)}},$$

the original problem becomes equivalent to (3.7). The Guo–Qi weighted shifted-ratio theorem yields (3.8), which is useful but not sufficiently strong to establish the required inequality.

In contrast, the two-variable Yang–Tian monotonicity theorem (Theorem 1) yields the stronger estimate (3.10). The elementary inequality (3.11) then gives the desired result.

Thus the tail-function and ratio-monotonicity methods serve different purposes. The tail-function approach provides an independent explicit lower bound and a self-contained partial proof, whereas the ratio-monotonicity approach yields the full positivity theorem. For this reason, the tail-function results should be retained even in the final version containing the complete proof. They constitute an independent quantitative contribution rather than merely an intermediate unsuccessful attempt.

4. Conclusions

We have investigated the inequality

$$F\left(z\right)=(z+2)\zeta (z+1)\zeta (z+3)-(z+1){\zeta }^2(z+2)-\zeta (z+1)\zeta (z+2)>0,z>0.$$

Our first approach is based on the tail function

$$\tau \left(s\right)=\zeta \left(s\right)-1.$$

Using a direct series decomposition together with the log-convex Turán-type inequality (2.1), we established the fundamental lower bound (3.3). This provides an independent quantitative lower bound and proves positivity for

$$z>z_0,$$

where $z_0$ is the unique positive solution of (3.4).

We then introduced the consecutive zeta ratio

$$r\left(x\right)={\displaystyle \frac{\zeta (x+1)}{\zeta \left(x\right)}}$$

and showed that the original problem is equivalent to

$$r(x+1)>{\displaystyle \frac{1+xr\left(x\right)}{x+1}},x>1.$$

Equivalently, defining

$$H\left(x\right)=x(1-r(x\left)\right),$$

the desired inequality is precisely the unit-step condition

$$H\left(x\right)>H(x+1).$$

The weighted shifted-ratio monotonicity theorem of Guo and Qi provides useful monotonicity information but is not by itself sufficient to establish this inequality. The stronger two-variable monotonicity theorem of Yang and Tian yields

$$r(x+1)>{\displaystyle \frac{2}{3-r\left(x\right)}}.$$

Since

$${\displaystyle \frac{2}{3-r\left(x\right)}}>{\displaystyle \frac{1+xr\left(x\right)}{x+1}},x>1,$$

the required consecutive-ratio inequality follows. Consequently,

$$(z+2)\zeta (z+1)\zeta (z+3)-(z+1){\zeta }^2(z+2)-\zeta (z+1)\zeta (z+2)>0$$

for every $z>0$.

The two approaches are complementary. The tail-function method provides an independent explicit lower bound, while the zeta-ratio monotonicity method supplies the stronger estimate required for positivity throughout the entire positive real axis. Together, these results connect the original problem with log-convexity, Turán-type inequalities, and contemporary monotonicity theory for ratios of shifted Riemann zeta values.