I want to show that
$$ 2\Phi(x)^2 - \Phi(2x) \geq 0 $$ for all $x<0$, where $\Phi$ is a cumulative distribution function of $N(0,1)$.
I am pretty sure that this holds since I've checked it with numerous $x$'s, but I am struggling to prove it.
I tried differentiating it, but it seems not working at least within my abilities. Could somebody help me with it? Thank you.
For all $x<0$, $$\Phi(x):=\int_{-\infty}^x\frac{\mathrm e^{-\frac{t^2}2}}{\sqrt{2\pi}}\,\mathrm dt\le\int_{-\infty}^x\frac{t\,\mathrm e^{-\frac{t^2}2}}{x\sqrt{2\pi}}\,\mathrm dt=-\frac{\Phi'(x)}{x}.$$ Since $\Phi''(x)=-x\Phi'(x)$, this implies the functional inequality $\Phi''\Phi\le\Phi'^2$ on $(-\infty,0]$, meaning that $\log\Phi$ is concave on this interval (we have $(\log\Phi)''=\frac{\Phi''\Phi-\Phi'^2}{\Phi^2}\le0$ there). Thus, for $x<0$, $$\log\Phi(x)=\log\Phi\!\left(\frac12\cdot2x+\frac12\cdot0\right)\ge\frac12\,\log\Phi(2x)+\frac12\,\log\Phi(0)=\frac12\,\log\frac{\Phi(2x)}2.$$ Multiplying by $2$ and taking the exponential gives $2\Phi(x)^2-\Phi(2x)\ge0$.