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.
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} $.