A normal distribution inequality 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.
$$f(x):= (x^2+1)N + xn-(xN+n)^2 > N^2$$
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,
$$f(x):=\int_0^\infty t^2n(t+x)dt-\Big(\int_0^\infty t\,n(t+x)dt\Big)^2>\Big(\int_0^\infty n(t-x)dt\Big)^2.$$
The form of the inequality is probably more transparent if we set $m=1-N$ and the inequality is equivalent to 
$$g(x)\equiv m[(x^2+1)(1-m)+2xn]-n(x+n) > 0.$$

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

Proof: $h$ 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?

$g(x)>0, \forall x\ge\sqrt{\frac{4}{3}}$

Proof:
\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, $g(x)>0, \forall x\ge x_0$.
 A: Here is a complete solution. The idea is to kill the entries of $N$ in two steps, by applying two appropriately constructed first-order differential operators, which will result in a simple elementary expression:
Let $b:=f-N^2$. As noted by cardinal, $b$ is an even function. So, it is enough to show that $b>0$ on $[0,\infty)$. Let 
$$
b_0(x):=\frac{b(x)}{x^2+1}
$$
and 
$$
b_1(x)=\pi\,  \left(x^2+1\right)^2 e^{x^2/2}\, b_0'(x). 
$$
Then $b_1'(x)=-e^{-\frac{x^2}{2}} \left(x^2+1\right)<0$, so that $b_1$ is decreasing. Also, $b_1(0)=0$. Hence, $b_1(x)<0$ for $x>0$, and so, $b_0$ is decreasing on $[0,\infty)$. Moreover, $b_0(x)\to0$ as $x\to\infty$. So, $b_0>0$ and hence $b>0$. 
A: Yes, the conjectured lower bound is true and can be proved using fairly simple, if somewhat tedious, analysis of derivatives.
First define
$$
b := f - N^2 = x(xN + n) - (xN + n)^2 + N(1-N)\>.
$$
The plan is to show that $b$ is a decreasing function bounded below by zero.
Let $u := x N + n$, so that $b = (x-u)u + (1-N)N = (x-u)u + (1-u')u'$. Since $u(-x) = -(x-u(x))$ and $N(-x) = 1-N(x)$, $b$ is an even function and so we restrict ourselves to the case $x \geq 0$.
Observe that $u' = N$, $u'' = n$, and $b(0) = (1/4) - (1/2\pi) > 0$.
By using the classical inequalities, valid for $x > 0$,
$$
\frac{xn}{x^2+1} \leq 1-N \leq \frac{n}{x} \>,
$$
on $(x-u)u$, it is straightforward to verify that $\lim_{x\to\infty} b(x) = 0$.
Now, using the fact $u = x u' + u''$,
$$
b' = 2u(1-u') - 2 u' u'' = 2 u' u''\left(\frac{(1-u')u}{u'u''} - 1\right) \>.
$$
So, if we can show that $\frac{(1-u')u}{u'u''} \leq 1$, we will be done. Plugging in the definitions yields $\frac{(1-u')u}{u'u''} = \frac{1-N}{n}(x+n/N)$.
Lemma 1. For $x \geq 0$, $n/N \leq a e^{-a x}$ where $a = \sqrt{2/\pi}$.
Proof. Define $g := a^{-1} e^{ax} n - N$. Then $g(0) = 0$ and
$$
g' = (1-x/a - e^{-ax})e^{ax} n < 0 \>.
$$
In particular, we have, $x+n/N \leq x + a e^{-a x}$ for any $x \geq 0$.
Lemma 2. For $x \geq 0$, $(1-N)/n \leq (x+a e^{-ax})^{-1}$.
Proof. Set $g := (x+ae^{-ax})^{-1} n - (1-N)$. Then, $g(0) = 0$ and
$$
g' = (a+ae^{-ax} + x - a^{-1} e^{ax}) \frac{a e^{-ax} n}{(x+a e^{-ax})^2}\>.
$$
The fraction on the right is positive, so we concentrate on the first term on the right. Let $z := a + a e^{-ax} + x - a^{-1} e^{ax}$. Then $z(0) = 2a - 1/a > 0$ and $\lim_{x\to\infty} z(x) = -\infty$. Furthermore,
$$
z' = - a^2 e^{-ax} + 1 - e^{ax} < 0 \>.
$$
Hence, $g'$ is positive for small $x$ and negative for large $x$. Since $\lim_{x\to\infty} g(x) = 0$, we conclude that $g \geq 0$.
This allows us to complete the proof, since by applying Lemma 1 and then Lemma 2, we have
$$
\frac{1-N}{n} (x + n/N) \leq \frac{1-N}{n} (x+a e^{-ax}) \leq 1 \>.
$$
Hence, $b' < 0$, so $b > 0$ as desired.
A: We may see that the inequality is true for every $|x|<0.597$ in the following way:
For a given value of $x$ consider the values of $N$ and $n$. The inequality will 
be true for this $x$ if the quadratic polynomial in $y$ 
$$(y^2+1)N+y\, n-(y N+n)^2-N^2$$
is always positive.  In other words the inequality is true for this $x$ as soon as 
the discriminant $\Delta$ of this quadratic is negative (the coefficient of 
$y^2$ being positive).
The discriminant is  $\Delta =n^2-4N^2(1-N)^2$.
Since $n^2<1/(2\pi)$, the inequality will be true for every $x$ such that
$4N^2(1-N)^2>1/(2\pi)$.  
Thus the inequality is true for every $x$ such that  $0.275214<N<0.724786$. 
This corresponds to the condition $|x|<0.597$. 
A: This is just to make explicit the functions in Iosif Pinelis' ingenious answer for posterity.
$$b_0(x):=\frac{b(x)}{x^2+1}=-N^2+\Big(1-\frac{2xn}{x^2+1}\Big)N+\frac{x-n}{x^2+1}n,$$
and
$$b_0'(x)=\frac{2n}{(x^2+1)^2}\,(-2N+1+xn).$$
$$b_1(x):=\frac{(x^2+1)^2}{2n}b_0'(x),$$
implying
$$b_1'(x)=-(1+x^2)n>0.$$
A: The Maple command $$asympt((x^2+1)*N(x)+x*n(x)-(x*N(x)+n(x))^2-N(x)^2, x, 8)$$ produces
$$ \left( {\frac {\sqrt {2}}{\sqrt {\pi }{x}^{3}}}-6\,{\frac {\sqrt {2}}
{\sqrt {\pi }{x}^{5}}}+O \left( {x}^{-7} \right)  \right) {\frac {1}{
\sqrt {{{\rm e}^{{x}^{2}}}}}}.
$$ Thus the inequality is true for big positive $x$. I leave the investigation of it on the finite interval on your own. The above asymptotics can be obtained by hand too. 
A: In view of 
$$f(x):= (x^2+1)N(x)+xn(x)-(x+N(x))^2-N(x)^2=$$ $$\left( {x}^{2}+1 \right)  \left( 1/2+1/2\,
{{\rm erf}\left(1/2\,\sqrt {2}x\right)} \right) +1/2\,{\frac {{{\rm e}
^{-1/2\,{x}^{2}}}\sqrt {2}x}{\sqrt {\pi }}}- $$ $$\left( x+1/2+1/2\,
{{\rm erf}\left(1/2\,\sqrt {2}x\right)} \right) ^{2}- \left( 1/2+1/2\,
{{\rm erf}\left(1/2\,\sqrt {2}x\right)} \right) ^{2}
 $$
 and its taylor expansion at $x=0$ 
 $$f(x)=-x+ \left( -1/2\,{\pi }^{-1}+{\frac {1}{2}}- \left( 1+1/2\,{\frac {
\sqrt {2}}{\sqrt {\pi }}} \right) ^{2} \right) {x}^{2}+O \left( {x}^{3
} \right) $$
the inequality under consideration seems to fail for small  positive values of $x$.
