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The following problem arised in my research work and has been challenging me for several days:

Prove that for all $r\in (0,1)$ and for all $x>0$, we have $$\frac{\sinh(r(2-r)x)}{r(2-r)x} \bigg[ \frac{\sinh((1-r)x)}{(1-r)x} \bigg]^2 > \frac{\sinh(x)}{x} \frac{\sinh((1-r)^2x)}{(1-r)^2x}.$$

Below are my thoughts. The inequality is obviously true when $x$ is large enough, thanks to the equivalence $\sinh(y)\sim \frac{1}{2}\exp(y)$, $y\rightarrow +\infty$, and to the fact that $$r(2-r)+2(1-r) > 1 + (1-r)^2$$ for $r\in(0,1)$. The inequality also holds true when $x$ is small enough, since it can be readily shown that the Taylor expansion of the difference between the LHS and the RHS has $$\frac{1}{45}r^2(1-r)^2(2-r)^2 x^4$$ as the leading term. However, the difficulty lies in establishing the claim for all $x>0$, which is "testified" by various graphic display softwares (Maple, Gnuplot, WolframAlpha).

A first natural transformation is to take the logarithm of both sides and to invoke the convexity of the function $x\mapsto\ln(\sinh(x)/x)$. But this leads nowhere: Jensen's inequality alone cannot help us deriving $$f(r(2-r)x)+2f((1-r)x) > f(x)+f((1-r)^2x)$$ for any convex function $f$ (consider a linear function and its opposite).

As a second attack, we can multiply both sides by $4r(2-r)(1-r)^2x^3$ and use the linearization formulae $$2\sinh(a)\sinh(b) = \cosh(a+b)-\cosh(a-b)$$ $$2\sinh(a)\cosh(b) = \sinh(a+b) + \sinh(a-b)$$ in order to obtain the equivalent inequality $$\sinh((2-r^2)x) -\sinh((r^2-4r+2)x) - 2\sinh(r(2-r)x) > 2r(2-r)x \big( \cosh((r^2-2r+2)x)- \cosh(r(2-r)x)\big).$$ The direct study of the difference between the new LHS and the new RHS appears to be unwieldy. Nevertheless, it is now straightforward to compute the Taylor series of this new difference. This series (whose radius of convergence is infinite) is equal to $$\sum_{n=0}^{\infty} \frac{a_n(r)}{(2n+1)!} x^{2n+1},$$ with $$a_n(r)= (2-r^2)^{2n+1} - (r^2-4r+2)^{2n+1} + 4n (r(2-r))^{2n+1} - (4n+2) r(2-r)(r^2-2r+2)^{2n}.$$ The first three coefficients are easily shown to vanish, i.e., $a_0(r)=a_1(r) = a_2(r) = 0$. Pushing further the calculations yields $$a_3(r)=448 r^3(1-r)^4(2-r)^3$$ $$a_4(r)=768 r^3(1-r)^4(2-r)^3 [3(1-r)^4 + 2(1-r)^2 + 3]$$ $$a_5(r)= 1408 r^3(1-r)^4(2-r)^3 [5(1-r)^8 + 12(1-r)^6 + 6(1-r)^4 + 12(1-r)^2 + 5].$$ These suggest that $a_n(r)$ is the product of $r^3(1-r)^4(2-r)^3$ with some even and symmetric polynomials in $1-r$ whose coefficients are all non-negative. The conjecture $a_n(r)> 0$ for $r\in(0,1)$ implies the desired result, but again I cannot prove it for $n\geq 6$.

Is there a general technique to prove that the power series coefficients of a given function are all non-negative? According to what I could find in the litterature, this seems to be a delicate issue.

Is there any fresh approach to the initial question? Many thanks for any help.

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    $\begingroup$ Where did you find this inequality, by the way? $\endgroup$ – Yuriy S Aug 7 '16 at 18:54
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    $\begingroup$ This is a long story. I study the Schrödinger equation $$-\frac{1}{2}\psi^{''} - Z_1 \delta_{X_1} \psi - Z_2 \delta_{X_2}\psi = E \psi$$ for a system consisting of one electron (whose wavefunction is $\psi$ and whose energy is $E$) and two nuclei of charges $Z_1$, $Z_2$ located at positions $X_1$, $X_2 \in [0,L]$ in a 1-D periodic domain of length $L$. [to be continued in next comment] $\endgroup$ – QHT Aug 8 '16 at 8:42
  • $\begingroup$ The symbol $\delta_X$ stands for the Dirac potential (which is the 1-D counterpart of the 3-D Coulomb potential). In this setting, it can be shown that all solutions $(\psi,E)$ with negative energy level (bound states) are given by $$E=-\frac{\zeta^2}{2}$$ and $$\psi(x)= \psi(X_1)\frac{Z_1}{\zeta} f(x-X_1;\zeta,L) + \psi(X_2) \frac{Z_2}{\zeta} f(x-X_2;\zeta,L),$$ where $$f(y; \zeta,L) = \frac{\cosh(\zeta(|y|_L-L/2))}{\sinh(\zeta L/2)}$$ and $| \cdot |_L$ the periodized absolute value (which coincides with the usual absolute value on $[-L/2,L/2]$). $\endgroup$ – QHT Aug 8 '16 at 9:13
  • $\begingroup$ Thank you, QHT. I didn't expect so much information, but it's nice to know that the problem comes from quantum mechanics, which is my favorite field $\endgroup$ – Yuriy S Aug 8 '16 at 9:16
  • $\begingroup$ To make the story short, the inequality I asked is sufficient to prove that under some condition on the data $Z_1$, $Z_2$, $|X_2-X_1|$, $L$, there is a unique bound-state solution. Indeed, the parameter $\zeta$ must satisfy two relations of compatibility obtained by specifying $x=X_1$ and $x=X_2$ in $\psi(x)$. For a nontrivial solution to exist, some $\zeta$-dependent $2\times 2$ matrix must have 1 as an eigenvalue, which yields a necessary condition for $\zeta$. $\endgroup$ – QHT Aug 8 '16 at 10:56
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First, let's replace your $r$ with $y=1-r$ $\qquad - \quad$ this makes the arguments a little nicer:

$$\frac{\sinh((1-y^2)x)}{(1-y^2)x} \left[ \frac{\sinh(yx)}{yx} \right]^2 > \frac{\sinh(x)}{x} \frac{\sinh(y^2x)}{y^2x}$$

Now we have the imaginary $\operatorname{sinc}$ functions here, and it makes sense to use the infinite product formula instead of the series:

$$\frac{\sinh(\pi z)}{\pi z}=\prod_{n=1}^\infty \left(1+\frac{z^2}{n^2} \right)$$

Let's set $x=\pi z$ then our inequality will become:

$$\prod_{n=1}^\infty \left(1+\frac{(1-y^2)^2z^2}{n^2} \right) \prod_{n=1}^\infty\left(1+\frac{y^2z^2}{n^2} \right)^2>\prod_{n=1}^\infty \left(1+\frac{z^2}{n^2} \right) \prod_{n=1}^\infty\left(1+\frac{y^4z^2}{n^2} \right)$$

All of the products converge, thus we can multiply them by term:

$$\prod_{n=1}^\infty \left(1+\frac{(1+y^4)z^2}{n^2}+\frac{y^2(2-3y^2+2y^4)z^4}{n^4}+\frac{y^4(1-2y^2+y^4)z^6}{n^6} \right)> \\ > \prod_{n=1}^\infty \left(1+\frac{(1+y^4)z^2}{n^2}+\frac{y^4z^4}{n^4}\right)$$

Now it is enough to prove that each term on the left is larger than corresponding term on the right. Then the product on the left will be larger as well:

$$1+\frac{(1+y^4)z^2}{n^2}+\frac{y^2(2-3y^2+2y^4)z^4}{n^4}+\frac{y^4(1-2y^2+y^4)z^6}{n^6} > 1+\frac{(1+y^4)z^2}{n^2}+\frac{y^4z^4}{n^4}$$

$$\frac{y^2(2-3y^2+2y^4)z^4}{n^4}+\frac{y^4(1-2y^2+y^4)z^6}{n^6} > \frac{y^4z^4}{n^4}$$

$$\frac{2y^2(1-2y^2+y^4)z^4}{n^4}+\frac{y^4(1-2y^2+y^4)z^6}{n^6} > 0$$

$$\frac{2y^2(1-y^2)^2z^4}{n^4}+\frac{y^4(1-y^2)^2z^6}{n^6} > 0$$

The last inequality is obviously true, thus we can trace all of the steps back and prove the original inequality.

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  • $\begingroup$ Great! The infinite product makes things much easier. Thanks very much. $\endgroup$ – QHT Aug 8 '16 at 8:22
  • $\begingroup$ I noticed I didn't use the fact that $y<1$ anywhere, so the inequality should probably work for general $y$ $\endgroup$ – Yuriy S Aug 8 '16 at 9:29
  • $\begingroup$ Interesting. In my problem, $r$ represents $2|X_2-X_1|_L/L$ (the ratio of the periodized distance between the nuclei over the half-period) and will always remain in $[0,1]$. As for the variable $x$ in the statement of the inequality, it represents the dimensionless quantity $\zeta L/2$. $\endgroup$ – QHT Aug 8 '16 at 11:13
  • $\begingroup$ @You're In My Eye Great lesson learnt on the use of product series $\endgroup$ – Laurent Duval Aug 8 '16 at 14:43

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