<s>[INST] Is this a lower bound for this integral? Let $T$ be the triangular region in the $xy$ plane with vertices $(-2,0) (2,0) $ and $(0.2)$. Then is this inequality true? $$\displaystyle\int\int_T e^{-(x^2+y^2)}dA \geq \pi\left(1-\dfrac{1}{\sqrt{e}}\right)$$
I can inscribe the half disk $D=\{(x,y)\in\mathbb{R}^2$ | $x^2+y^2=2, y\geq 0\}$ inside $T$. Then it is clear that,
$$ \displaystyle\int\int_Te^{-(x^2+y^2)}dA \geq \displaystyle\int\int_De^{-(x^2+y^2)}dA$$
Changing to polar coordinates I get the RHS as $\dfrac{\pi}{2}\left(1-\dfrac{1}{e^2}\right)$
Clearly, $$\dfrac{\pi}{2}\left(1-\dfrac{1}{e^2}\right)\geq \dfrac{\pi}{2}\left(1-\dfrac{1}{\sqrt{e}}\right)$$
So, $$\displaystyle\int\int_T e^{-(x^2+y^2)}dA \geq\dfrac{\pi}{2}\left(1-\dfrac{1}{\sqrt{e}}\right)$$
Can I get the RHS without the $\dfrac{1}{2}$?
 [/INST] Twice your integral is the integral over the square with vertices at $(\pm 2,0)$ and $(0,\pm 2)$, hence:
$$2I = \int_{-\sqrt{2}}^{+\sqrt{2}}\int_{-\sqrt{2}}^{+\sqrt{2}} e^{-(x^2+y^2)}\,dx\,dy =\pi\cdot\text{Erf}(\sqrt{2})^2\tag{1}$$
but due to the continued fraction representation for the error function,
$$ \text{Erf}(\sqrt{2})\geq 1-\frac{1}{e^2\sqrt{2\pi}}\tag{2} $$
hence:
$$ I \geq \frac{\pi}{2}\left(1-\frac{2}{e^2\sqrt{2\pi}}\right)>\frac{7}{5}\tag{3}$$
but since:
$$ \text{Erf}(\sqrt{2}) \leq 1-\frac{4\sqrt{2}-2}{7e^2\sqrt{\pi}}\tag{4}$$
we also have:
$$ I \leq \frac{\left(2-4 \sqrt{2}+7 e^2 \sqrt{\pi }\right)^2}{98 e^4}<\frac{42}{29}\tag{5}$$
hence your initial inequality does not hold.
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