How to prove that second derivative of $\log\big(\int_{-\infty}^x e^{\frac{-t^2}2} dt\big)$ is $>-1$? Let $\Phi(x)=\int_{-\infty}^x e^{\frac{-t^2}2} dt$.
How can I prove that
$$\left[\frac{e^{\frac{-x^2}2}}{\Phi(x)}\right]'>-1?$$
I could prove that its $lim$ at $-\infty$ is $-1$ and at $\infty$ it is $0$ but the l'Hospital rule did not seem to be enough to prove the inequality. At zero the value is $-\frac 2\pi$.

Let $\phi(x)=e^\frac{-x^2}{2}$. Note that $\phi'(x)=-x\phi(x)$.
$$x\sqrt{x^2+4}+x^2+2>0\\
\iff x+\sqrt{x^2+4}>\frac{2}{\sqrt{x^2+4}}\\
\iff \frac{1}{2}\left(-x+\sqrt{x^2+4}\right)>-x+\frac{1}{\sqrt{x^2+4}}\\
\iff 1>\frac{1}{2}\left(x+\sqrt{x^2+4}\right)\left(-x+\frac{1}{\sqrt{x^2+4}}\right)\\
\iff \phi(x)>\frac{1}{2}\phi(x)\left(x+\sqrt{x^2+4}\right)\left(-x+\frac{1}{\sqrt{x^2+4}}\right)\\
\overset{\text{integrating}}{\implies} \Phi(x)>\frac{1}{2}\phi(x)\left(x+\sqrt{x^2+4}\right)\\
\iff \left(\Phi(x)-\frac{1}{2}x\phi(x)\right)^2>\phi(x)^2(1+\frac{x^2}{4})\\
\implies x\phi(x)\Phi(x)+\phi(x)^2<\Phi(x)^2\\
\iff \left(\phi(x)/\Phi(x)\right)'>-1.$$
