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I am trying to prove an inequality involving exponentials, namely that for all $x,y>0$, \begin{equation} \big(e^{x^2}-1\big)\big(e^{y^2}-1\big) \geq \big(e^{xy}-1\big)^2 \end{equation} Any suggestions would be much appreciated.

Update: I tried moving everything to one side, expanding and looking at first-order derivatives in hopes of observing monotonicity, as well as rewriting the inequality as \begin{equation} \frac{e^{xy\,\cdot\,\tfrac{x}{y}}-1}{e^{xy}-1} \geq \frac{e^{y^2\,\cdot\,\tfrac{x}{y}}-1}{e^{y^2}-1} \end{equation} assuming $x/y>1$ constant and looking again at the derivative, but was unsuccessful.

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  • $\begingroup$ I solved your problem. If you want to see my solution, show please your attempts. $\endgroup$ – Michael Rozenberg Sep 22 at 19:12
  • $\begingroup$ Thanks, Michael. I updated my original question. I mostly tried rewriting the inequality in functional form and looking at partial derivatives, but was unsuccessful. By the way, a solution would be even better than a suggestion. $\endgroup$ – Andrei Sep 22 at 19:16
  • $\begingroup$ Show, haw exactly you made this expanding. I a posting my solution already... $\endgroup$ – Michael Rozenberg Sep 22 at 19:17
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By C-S $$(e^{x^2}-1)(e^{y^2}-1)=\left(x^2+\frac{x^4}{2!}+\frac{x^6}{3!}+...\right)\left(y^2+\frac{y^4}{2!}+\frac{y^6}{3!}+...\right)\geq$$ $$\geq\left(xy+\frac{x^2y^2}{2!}+\frac{x^3y^3}{3!}+...\right)^2=\left(e^{xy}-1\right)^2.$$

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  • $\begingroup$ Thank you. I assume taking limits after using the Cauchy-Schwarz inequality is implied. $\endgroup$ – Andrei Sep 22 at 19:29
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    $\begingroup$ @Andrei Yes, but we can use C-S also in the dimension infinity. Just our series converges. $\endgroup$ – Michael Rozenberg Sep 22 at 19:34
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    $\begingroup$ @Andrei Cauchy-Schwarz is valid in Hilbert space. $\endgroup$ – Jean Marie Sep 23 at 2:19
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An alternative approach: With the substitutions $x=e^{u/2}$, $y = e^{v/2}$ and taking logarithms, the inequality becomes $$ 2 \log (e^{\large e^{(u+v)/2}}-1) \le \log (e^{\large e^u}-1) + \log (e^{\large e^v}-1) $$ so that is remains to show that the function $$ f(u) = \log (e^{\large e^u}-1) $$ is convex. A straightforward calculation gives $$ f''(u) = \frac{e^{\large u+e^u} (e^{\large e^u}-e^u-1)}{(e^{\large e^u}-1)^2} $$ and that is positive because $e^x > 1+x$ for all positive $x$.

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  • $\begingroup$ Nice proof and ingenious substitutions, thank you. $\endgroup$ – Andrei Sep 23 at 8:50
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Alternative solution:

Let $x^2 = u, y^2 = v$. The inequality is written as $$(\mathrm{e}^u - 1)(\mathrm{e}^v - 1) \ge (\mathrm{e}^{\sqrt{uv}}-1)^2.$$ Let $uv = a > 0$ be fixed. Let $$f(u) \triangleq (\mathrm{e}^u - 1)(\mathrm{e}^{a/u} - 1) - (\mathrm{e}^{\sqrt{a}}-1)^2.$$ We have \begin{align} f'(u) &= \mathrm{e}^u (\mathrm{e}^{a/u} - 1) + (\mathrm{e}^u - 1) (-a/u^2)\mathrm{e}^{a/u}\\ &= \mathrm{e}^u \mathrm{e}^{a/u} \frac{a}{u} \left(\frac{1 - \mathrm{e}^{-a/u}}{a/u} - \frac{1 - \mathrm{e}^{-u}}{u}\right). \end{align} It is easy to prove that $y\mapsto \frac{1 - \mathrm{e}^{-y}}{y}$ is strictly decreasing on $(0, \infty)$ (simply taking derivative). Thus, $f'(u) < 0$ on $(0, \sqrt{a})$, $f'(u) > 0$ on $(\sqrt{a}, \infty)$, and $f'(\sqrt{a}) = 0$. Thus, $f(u) \ge f(\sqrt{a}) = 0$. We are done.

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  • $\begingroup$ Definitely better than my deleted answer . (+1) $\endgroup$ – Erik Satie Sep 24 at 10:43
  • $\begingroup$ @ErikSatie Your idea is similar and works. $\endgroup$ – River Li Sep 24 at 11:55
  • $\begingroup$ Yes but you are the first so... $\endgroup$ – Erik Satie Sep 24 at 13:16
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Assuming wlog $x\ge y$ we have

$$\big(e^{x^2}-1\big)\big(e^{y^2}-1\big) \ge \big(e^{xy}-1\big)^2 \iff \frac{e^{x^2}-1}{e^{xy}-1}\ge \frac{e^{xy}-1}{e^{y^2}-1}$$

then we reduce to prove that for $a=\frac x y\ge 1$ and $u>1$

$$f(u)=\frac{u^a-1}{u-1}$$

is increasing, which is true indeed

$$f'(u)=\frac{(a-1)u^{a}-au^{a-1}+1}{(u-1)^2}\ge 0$$

since

$$g(u) =(a-1)u^{a}-au^{a-1}+1\implies g(1)=0$$

and

$$g'(u)=a(a-1)u^{a-1}-a(a-1)u^{a-2}=a(a-1)u^{a-1}\left(1-\frac1u\right)\ge 0$$

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    $\begingroup$ Thank you, I had tried a similar approach but got stuck at proving that the first derivative is positive. $\endgroup$ – Andrei Sep 23 at 8:51
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    $\begingroup$ @Andrei You are welcome! A great simplification is assume $u=e^x>1$ as variable. $\endgroup$ – user Sep 23 at 8:53

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