Standard normal distribution inequality I want to know how to prove the following inequality that seems to be true numerically.
Let $n(x)$ be the density of the standard normal, and $N(x)$ be the cdf of standard normal. Then, for $x\geq 0$,
$\left(\frac{n(x)}{1-N(x)}-x \right)\left(\frac{2n(x)}{1-N(x)}-x\right)\geq 1$
Thanks, and sorry that I don't know how to write the math symbols in this environment.
 A: I get that
$\left(\frac{n(x)}{M(x)}-x \right)\left(\frac{2n(x)}{M(x)}-x\right)
- 1
\approx \frac{3}{x^2}
$
for large $x$.
Since
$N(x)
=\int_{-\infty}^x n(t)dt
$,
$1-N(x)
=\int_x^{\infty} n(t)dt
$.
Call this $M(x)$.
The inequality becomes
$\left(\frac{n(x)}{M(x)}-x \right)\left(\frac{2n(x)}{M(x)}-x\right)\geq 1
$.
Multiplying by
$M^2(x)$,
I get
$(n(x) - xM(x))(2n(x) - xM(x))
\ge M^2(x)
$
or
$2n^2(x) -3xn(x)M(x)+x^2M^2(x)
\ge M^2(x)
$
or
$2n^2(x) -3xn(x)M(x)+(x^2-1)M^2(x)
\ge 0
$.
The purported inequality above
thus becomes
$P(x)
=2n^2(x) -3xn(x)M(x)+(x^2-1)M^2(x)
\ge 0
$.
Since
$M(0) = 1/2$
and
$n(0)
=1/\sqrt{2\pi}
\approx 0.4
$,
$P(0)
\approx 0.32-1/4
= 0.07
> 0
$.
Let's see what happens
for large $x$.
Asymptotically,
$M(x)
\approx n(x)/x
$.
Therefore,
for large $x$,
$\begin{array}\\
P(x)
&= 2n^2(x) -3xn(x)M(x)+(x^2-1)M^2(x)\\
&\approx 2n^2(x) -3xn(x)(n(x)/x)+(x^2-1)(n(x)/x)^2\\
&= 2n^2(x) -3n^2(x)+(1-1/x^2)n^2(x)\\
&= -n^2(x)/x^2\\
\end{array}
$
Therefore,
modulo errors on my part,
$P(x)
\approx -n^2(x)/x^2
$,
so it is 
negative
and small.
Note:
A comment said that this was
not correct,
so I'll take an
additional term in the
expansion and see what happens.
Asymptotically,
$M(x)
\approx \frac{n(x)}{x}(1-\frac1{x^2})
$.
Therefore,
for large $x$,
$\begin{array}\\
P(x)
&= 2n^2(x) -3xn(x)M(x)+(x^2-1)M^2(x)\\
&\approx 2n^2(x) -3xn(x)(n(x)/x)(1-1/x^2)+(x^2-1)(n(x)/x)^2(1-1/x^2)^2\\
&= 2n^2(x) -3n^2(x)(1-1/x^2)+n^2(x)(1-1/x^2)^3\\
&= 2n^2(x) -3n^2(x)+3n^2(x)/x^2+n^2(x)(1-3/x^2+3/x^4-1/x^6)\\
&= -n^2(x)+3n^2(x)/x^2+n^2(x)-n^2(x)(3/x^2-3/x^4+1/x^6)\\
&= n^2(x)(3/x^4-1/x^6)\\
&\approx 3n^2(x)/x^4\\
\end{array}
$
Therefore,
modulo errors on my part,
$P(x)
\approx 3n^2(x)/x^4
$,
so it is 
positive
and small.
Looks like the comment
was correct.
Since I multiplied by $M^2(x)$
to get this inequality,
the difference has to be
divided by $M^2(x)$,
so it is
$\frac{3n^2(x)/x^4}{M^2(x)}
\approx \frac{3n^2(x)/x^4}{(n(x)/x)^2}
=\frac{3}{x^2}
$.
