A question on an integral inequality I happened to have learned the following question:
Let $K \subset \mathbb{R}^2$ be a convex domain with $0\in K$. Prove that if
$$\frac{1}{2\pi}\int_{K}e^{-\frac{|x|^2}{2}}\, dx=\frac{1}{2},$$
then
$$\frac{1}{2\pi}\int_{K}e^{-\frac{|x|^2}{2}}|x|^2\, dx<1.$$

Numerical results indicate that if $K$ is a disk centered at origin, then if the first integral is equal to $1/2$, then the second integral is even less than $1/2$. For aibitrary convex set $K$, I'm thinking how to use this numerical fact to prove the similar thing, but I haven't had any clue yet so far. Any comments or ideas would be really appreciated.
 A: First, we prove the case $K \subset \Bbb R^2$, and in the end of this answer, we generalize the result for  $K \subset \Bbb R^n$

It suffices to prove that
$$\frac{1}{2\pi}\iint_{K}e^{-\frac{|x|^2}{2}}|x|^2dx<2\frac{1}{2\pi}\iint_{K}e^{-\frac{|x|^2}{2}}dx = 1$$
or
$$L:=\frac{1}{2\pi}\iint_{K}e^{-\frac{|x|^2}{2}}(2-|x|^2)dx >0  \quad ,K \subset\Bbb R^2   \tag{1}$$
Make a change of variable to Polar Coordinates
\begin{cases}
x_1 = r\cos(\theta)\\
x_2 = r\sin(\theta)\\
\end{cases}
Because $0 \in K \subset \Bbb R^2  $ and $K$ is convex then the region of integration is
\begin{cases}
\theta \in (0,2\pi)\\
r \in (0,r_{\max}(\theta))\\
\end{cases}
with $r_{\max}(\theta)$ is a positive function depending on $\theta$
From $(1)$, we have
\begin{align}
L &=\frac{1}{2\pi}\iint_{\theta \in (0,2\pi), r \in (0,r_{\max}(\theta))}e^{-\frac{r^2}{2}}(2-r^2)rdrd\theta \\
&=\frac{1}{2\pi}\int_0^{2\pi} \left(\int_0^{r_{\max}(\theta)}e^{-\frac{r^2}{2}}(2-r^2)rdr \right)d\theta \tag{2}
\end{align}
$(2)$ holds true if we can prove that
$$f(x):=\int_0^{x}e^{-\frac{r^2}{2}}(2-r^2)rdr > 0 \quad ,\forall x>0 \tag{3}$$
Let's study the derivative of $f$:  $f'(x) = \frac{1}{2\pi}e^{-\frac{x^2}{2}}(2-x^2)x$ and then we can deduce that the function $f(y)$ is increasing in $(0,\sqrt{2})$ and decreasing in $(\sqrt{2},+\infty)$. Hence
$$f(x)\ge \min\{f(0),f(+\infty)\}$$
with
\begin{align}
f(0) &=0\\
f(+\infty)&=\frac{1}{2\pi}\int_{0}^{+\infty}e^{-\frac{r^2}{2}}(2-r^2)rdr = 0
\end{align}
Then $f(y) \ge \min\{f(0),f(+\infty)\} = 0$ or $(3)$ holds true.
We deduce that $(2)$ holds true, and so $(1)$ holds true.
Conclusion:
$$\frac{1}{2\pi}\iint_{K}e^{-\frac{|x|^2}{2}}|x|^2dx< 1 $$
Q.E.D

Remark: by the same method, if $0 \in K \subset \Bbb R^n$ and $K$ is convex, we can have a more general result
$$\frac{1}{2\pi}\iint_{K}e^{-\frac{|x|^2}{2}}|x|^2dx<n\frac{1}{2\pi}\iint_{K}e^{-\frac{|x|^2}{2}}dx = \frac{n}{2}$$
by studying the function
$$f_n(x)= \int_0^{x}e^{-\frac{r^2}{2}}(n-r^2)r^{n-1}dr$$
