How to prove that $\forall a\in\mathbb R^+: \frac{\int_a^\infty x\cdot e^{-x^2/2}dx}{\int_a^\infty e^{-x^2/2}dx}\le a + \sqrt{2/\pi}$? Let $a\in\mathbb R^+$ and $X\sim\mathcal N(0,1)$.
I want to prove that
$\mathbb E[X | X\ge a] = \frac{\int_a^\infty x\cdot e^{-x^2/2}dx}{\int_a^\infty e^{-x^2/2}dx}\le a + \sqrt{2/\pi}$.
Any ideas/directions?

Playing with it in Wolfram Alpha, it seems that the inequality hold, with equality for $a=0$.
 A: A partial answer. Take $f(x) = \exp(-x^2/2)$.  Then for $x > 0$ $$f(x) = -\frac{f’(x)}x.$$
Take $a>1$ and use partial integration twice: $$\begin{eqnarray}\int_a^{\infty}f(x)\mathrm dx &=& -\int_a^{\infty}\frac{f’(x)}x\mathrm dx\\
 &=& -\left[\frac{f(x)}x\right]_a^{\infty} - \int_a^{\infty} \frac{f(x)}{x^2}\mathrm dx\\
&=& \frac{f(a)}a - \int_a^{\infty} \frac{f(x)}{x^2}\mathrm dx
\end{eqnarray}$$
$$\begin{eqnarray}\int_a^{\infty}\frac{f(x)}{x^2}\mathrm dx &=& -\int_a^{\infty}\frac{f’(x)}{x^3}\mathrm dx\\
 &=& -\left[\frac{f(x)}{x^3}\right]_a^{\infty} - 3 \int_a^{\infty} \frac{f(x)}{x^4}\mathrm dx\\
&=& \frac{f(a)}{a^3} - 3 \int_a^{\infty} \frac{f(x)}{x^4}\mathrm dx < \frac{f(a)}{a^3}
\end{eqnarray}$$
Combine these two results to conclude $$\int_a^{\infty}f(x)\mathrm dx > f(a)\left(\frac1a - \frac1{a^3}\right).$$
This suffices to show your result for big enough $a$ (such that $a/(a^2-1) \leq \sqrt{2/\pi}$.)
A: Here is a proof based on know asymptotics for $P[X>a]$ as $a\rightarrow\infty$ to establish the validity of the statement for large $a$, and then, using  differential Calculus to establish the validity of the statement for all $a\geq0$.
It is known that for  $1-F(a)=P[X>a]$ (See Feller, Introduction to Probability, Vol 1, 1968, section 7.1))
$$\Big(\frac1a-\frac1{a^3}\big)\phi(a)\leq 1-F(a)\leq \frac1a\phi(a)$$
where $\phi(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$. (This can be obtained by integrating the inequalities
$$(1-3x^{-4})\phi(x)\leq \phi(x)\leq (1+x^{-2})\phi(x)$$
over $[a,\infty)$.)
A simple substitution $u=x^2/2$ yields
$$\int^\infty_ax\phi(x)\,dx = \frac{1}{\sqrt{2\pi}}e^{-a^2/2}=\phi(a)$$
Hence, for $a>1$
$$
a\leq\frac{\phi(a)}{1-F(a)}\leq a+\frac{a}{a^2-1}
$$
Thus, the stamens is valid for all $a$ large enough.
Now consider the function
$$G(a)=\phi(a)-(a+c)\bar{F}(a),\qquad\text{where}\quad \bar{F}(a)=1-F(a),\quad c=\sqrt{2/\pi}$$
$G(0)=0$ as $c=2\phi(0)$, and $G(\infty):=\lim_{a\rightarrow\infty}G(a)=0$ as $\phi(a)\xrightarrow{a\rightarrow\infty}0$ and $a\bar{F}(a)\xrightarrow{a\rightarrow\infty}0$ (by Markov-Chebyshev's inequality for example).
$$\begin{align}
G'(a)&=c\phi(a)-\bar{F}(a)\\
G''(a)&=(1-ac)\phi(a)
\end{align}$$
There is only one inflection point, namely $a_*=c^{-1}=\sqrt{\pi/2}$; moreover, it follows easily that $G$ is convex on $(-\infty,a_*]$, and concave on $[a_*,\infty)$. Since $G'(0)=\frac{1}{\pi}-\frac12<0$, $G$ is strictly monotone decreasing in an interval containing $a=0$. This, along with the facts that  $G'(\infty)=0$ and that $G(a)<0$ for all $a$ large enough, implies that $G$ cannot take positive values on $[0,\infty)$ (otherwise there would be more than one inflection points) and so, $G(a)< 0$ for all $a>0$.
Here is a graph of $G$ in the interval $[0,5]$.

