A normal distribution inequality

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.

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 $b>0$.