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 $x,y>0$,
\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.
 A: By C-S $$(e^{x^2}-1)(e^{y^2}-1)=\left(x^2+\frac{x^4}{2!}+\frac{x^6}{3!}+...\right)\left(y^2+\frac{y^4}{2!}+\frac{y^6}{3!}+...\right)\geq$$
$$\geq\left(xy+\frac{x^2y^2}{2!}+\frac{x^3y^3}{3!}+...\right)^2=\left(e^{xy}-1\right)^2.$$
A: 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 $x$.
A: 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
$$g'(u)=a(a-1)u^{a-1}-a(a-1)u^{a-2}=a(a-1)u^{a-1}\left(1-\frac1u\right)\ge 0$$
A: 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.
