# How to show this function is increasing? Related to Normal distribution.

Numerically, it seems the following function $F(x)$ is increasing in $x$. How can I show it analytically?

$$F(x)=G(x)L'(x)$$

where $L(x)=\frac{(1-G(x))^3}{G'(x)}$ and $G(x)=\int_{-\infty}^x \frac{e^{- \frac{ 1}{2} y^2}}{\sqrt{2\pi}} dy$, and $L'(x)$ is derivative of $L(x)$. Note that $G(\cdot)$ can be interpreted as the CDF of standard normal distribution and $L(\cdot)$ is proportional to inverse hazard rate of standard normal distribution. We can show that $L(\cdot)$ is decreasing.

Let $\phi(x) =\frac{e^{- \frac{ 1}{2} x^2}}{\sqrt{2\pi}}$.
From $G(x) =\int_{-\infty}^x \phi(y) \, dy$, $G'(x)=\phi(x)$ and $1-G(x)=\int_x^{\infty} \phi(y) \, dy$.
Also, $$G(x) =1-\phi(x)(x+\frac{x^3}{3}+\frac{x^5}{3\cdot 5}+\frac{x^7}{3\cdot 5 \cdot 7} +\cdots) =1-\phi(x)(x+h(x))$$ so $1-G(x)=\phi(x)(x+h(x))$ and $L(x) =\frac{(1-G(x))^3}{G'(x)} =\frac{(\phi(x)(x+h(x)))^3}{\phi(x)} =\phi^2(x)(x+h(x))^3$.
Then $L'(x) =2\phi(x)\phi'(x)(x+h(x))^3 +3\phi^2(x)(x+h(x))^3(1+h'(x))$.