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

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

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 $x$.

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

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