Let $n(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}$, and $N(x) = \int_{-\infty}^x n(t)dt$. I have plotted the curves of the both sides of the following inequality. The graph shows that the following inequality may be true. $$f(x)\equiv (x^2+1)N + xn-(xN+n)^2 > N^2$$ where the dependency of $n$ and $N$ on $x$ are absorbed into the function symbols. However, I have not succeeded in providing a full proof except for $x$ above some positive number, with the help of various Mill's Ratio $\frac{m}{n}$ bounds.

I am asking for help in proving the above inequality or providing an $x$ that violates the above inequality. Judging from the aforementioned plot I am pretty confident the validity of the inequality, though.

The left hand side is actually the variance of a truncated normal distribution. I am trying to give it a lower bound. More explicitly, $$f(x)\equiv\int_0^\infty t^2n(t+x)dt-\Big(\int_0^\infty t\,n(t+x)dt\Big)^2>\Big(\int_0^\infty n(t-x)dt\Big)^2.$$

The form of the inequality is probably more transparent if we set $m=1-N$ and the inequality is equivalent to $$g(x)\equiv m[(x^2+1)(1-m)+2xn]-n(x+n) > 0.$$

Incidentally, I have proved that $N$ is the upper bound of $f$, i.e. $$(x^2+1)N + xn-(xN+n)^2 < N$$ or $$h(x)\equiv x^2 m(1-m)-n[x(1-2m)+n]<0$$ as follows.

$h$ is an even function and $h(0)<0$, so we only need to consider $x>0$. From the integration by part of $m(x)$ and dropping a negative term, we have $$xm<n, \forall x>0.$$ The first term of $h(x)$ is then bounded and \begin{eqnarray} h(x)&<&x(1-m)n-n[x(1-2m)+n] \\ &=& n(xm-n) \\ &<& 0, \end{eqnarray} where last inequality is obtained by using $xm<n$ again.

The lower bound of $f(x)$ appears to be more difficult since it requires tighter approximation of $m$ without singularity at $x=0$. I can prove the lower bound for $x$ greater than some positive number. I know I need to stitch the small and large regions of positive $x$ together, but I have not carried the detailed computation out yet. Does anyone have more clever trick to accomplish this task?

Here is the proof for $g(x)>0, \forall x\ge\sqrt{\frac{4}{3}}$. \begin{align} \frac{dg}{dx} &= 2n[xr(1-m)-2(0.5-m)] \\ &= 2n^2[(xr-1)n^{-1}+(2-xr)r] \end{align} where $r:=\frac{m}{n}$. In what follows we will use the first expression. The second expression is an alternative which I keep just for maybe future reference. Since $$r<\frac{1}{x}\Big(1-\frac{1}{x^2+3}\Big), \forall x>0,$$ \begin{align} \frac{dg}{dx} &< \frac{2n^2}{x^2+3}(-n^{-1}+(x^2+4)r) \\ &<\frac{2n^2}{x^2+3}\Big(-n^{-1}+x\Big(1+\frac{4}{x^2}\Big)\Big), \end{align} where on the last line we apply the $r$ bound again. Choose $x\ge x_0:=\sqrt{\frac{4}{3}}$, $$n^{-1}-x\Big(1+\frac{4}{x^2}\Big)>n^{-1}-4x.$$ It can be shown that $n^{-1}-4x$ is positive at $x=x_0$ and its derivative is always positive for $x\ge x_0$. We thus have $$\frac{dg}{dx}<0, \forall x\ge x_0.$$ It is easy to see that $g(x)>0$ for sufficiently large $x$. Therefore, $g(x)>0, \forall x\ge x_0$.