# How do I show $(e^{x^2}-1)(e^{y^2}-1) \geq (e^{xy}-1)^2$ for all $x,y> 0$?

I am trying to prove an inequality involving exponentials, namely that for all $$x,y>0$$, $$$$\big(e^{x^2}-1\big)\big(e^{y^2}-1\big) \geq \big(e^{xy}-1\big)^2$$$$ 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 $$$$\frac{e^{xy\,\cdot\,\tfrac{x}{y}}-1}{e^{xy}-1} \geq \frac{e^{y^2\,\cdot\,\tfrac{x}{y}}-1}{e^{y^2}-1}$$$$ assuming $$x/y>1$$ constant and looking again at the derivative, but was unsuccessful.

• I solved your problem. If you want to see my solution, show please your attempts. – Michael Rozenberg Sep 22 at 19:12
• 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. – Andrei Sep 22 at 19:16
• Show, haw exactly you made this expanding. I a posting my solution already... – Michael Rozenberg Sep 22 at 19:17

By C-S $$(e^{x^2}-1)(e^{y^2}-1)=\left(x^2+\frac{x^4}{2!}+\frac{x^6}{3!}+...\right)\left(y^2+\frac{y^4}{2!}+\frac{y^6}{3!}+...\right)\geq$$ $$\geq\left(xy+\frac{x^2y^2}{2!}+\frac{x^3y^3}{3!}+...\right)^2=\left(e^{xy}-1\right)^2.$$

• Thank you. I assume taking limits after using the Cauchy-Schwarz inequality is implied. – Andrei Sep 22 at 19:29
• @Andrei Yes, but we can use C-S also in the dimension infinity. Just our series converges. – Michael Rozenberg Sep 22 at 19:34
• @Andrei Cauchy-Schwarz is valid in Hilbert space. – Jean Marie Sep 23 at 2:19

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 $$x$$.

• Nice proof and ingenious substitutions, thank you. – Andrei Sep 23 at 8:50

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.

• Definitely better than my deleted answer . (+1) – Erik Satie Sep 24 at 10:43
• @ErikSatie Your idea is similar and works. – River Li Sep 24 at 11:55
• Yes but you are the first so... – Erik Satie Sep 24 at 13:16

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

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

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