<s>[INST] Prove $\iint_{x^{2}+y^{2} \leq 1} e^{x^{2}+y^{2}} d x d y \leq\left[\int_{-\frac{\sqrt{\pi}}{2}}^{\frac{\sqrt{\pi}}{2}} e^{x^{2}} d x\right]^{2} $ The formula we need to prove is
$$
\displaystyle \iint_{x^{2}+y^{2} \leq 1} e^{x^{2}+y^{2}} d x d y \leq\left[\int_{-\frac{\sqrt{\pi}}{2}}^{\frac{\sqrt{\pi}}{2}} e^{x^{2}} d x\right]^{2} 
$$
As I already done the left side
$$
\iint_{x^{2}+y^{2} \leq 1} e^{x^{2}+y^{2}} d x d y  = \int_{-\pi}^{\pi} \mathrm{d}\varphi \int_{0}^{1} e^{r^2}r\mathrm{d}r  = 2\pi \int_{0}^{1} e^{r^2}r\mathrm{d}r = \left.\pi e^{r^2}\right|_{0}^{1} = \pi(e-1)
$$
I still have no idea about the right side.
 [/INST] Integral on the left is the integration over the $\color{red}{red}$ circle, and the integral on the right is the integration over the $\color{blue}{blue}$ square (see the picture below).
We note that the area of the circle is $S_\circ = \pi\cdot 1^2 = \pi$ and the area of the square is $S_\square = (\sqrt \pi)^2 = \pi$. This means the $\color{green}{green}$ and $\color{orange}{orange}$ areas are equal too.
However, note that $$e^{\color{green}{x^2+y^2}} \le e^{\color{orange}{x^2+y^2}} $$ since $\color{green}{x^2+y^2} \le 1$ and $\color{orange}{x^2+y^2} \ge 1$ and therefore
$$\begin{align}
\iint\limits_{x^2+y^2\le 1} e^{x^2+y^2} dxdy 
&\le \iint\limits_{\text{square}} e^{x^2+y^2}dxdy \\
&= \int_{-\sqrt\pi/2}^{\sqrt\pi/2}e^{x^2}dx\int_{-\sqrt\pi/2}^{\sqrt\pi/2}e^{y^2}dy \\
&= \left(\int_{-\sqrt\pi/2}^{\sqrt\pi/2}e^{x^2}dx\right)^2
\end{align}$$

$\qquad\qquad\qquad$
</s>