Inequality for the maximum of the absolute value of two normal distributed random-variables I would like to show following statement:

For $M\geq 2,\ X_1,\dots,X_M\sim^{iid}\mathcal{N}(0,1)$ independent, it holds $P(\max_{i=1,\dots,M}\lvert X_i\rvert\geq y)\leq Me^{-y^2/2}$. 

I think it is possible to show it via induction, but I have got problems at the start:
So, I want to show $P\left(\max(\lvert X_1\rvert,\lvert X_2\rvert)\geq y \right)\leq 2 e^{-y^2/2}$. One can show $P\left(\max(\lvert X_1\rvert,\lvert X_2\rvert )\geq y\right)=4\Phi(y)(1-\Phi(y))$ where $\Phi$ denotes the distribution function. I know that $1-\Phi(y)\leq e^{-y^2/2}$, so this inequality seems to be a little sharper to me. Maybe someone knows how to prove this or an alternative prooving strategy.

Direct proof: Let $G(y)=1-4\Phi(y)(1-\Phi(y))=1+4\Phi(y)^2-4\Phi(y)$, the distribution function for $max(|X_1|,|X_2|)$.  The density function $g(y)=8\Phi(y)\phi(y)-4\phi(y)$.  Now $\Phi(y)=\frac{1}{2}+\frac{1}{\sqrt{2\pi}}\int_0^y e^{-\frac{u^3}{2}}du$,  so $g(y)=\frac{4}{\pi}e^{-\frac{y^2}{2}}\int_0^y e^{-\frac{u^2}{2}}du$.  Therefore $G(y)=\frac{4}{\pi}\int_0^y\int_0^ve^{-\frac{u^2+v^2}{2}}dudv$.  Switching to polar coordinates (note), where $0\le \theta \le \frac{\pi}{4}$, we get $G(y)\gt \int_0^yre^{-\frac{r^2}{2}}dr=1-e^{-\frac{y^2}{2}}$, or $P(max|X_1|,|X_2|)\ge y)\le e^{-\frac{y^2}{2}}$
note: The domain of the double integral is a right triangle in the $(u,v)$ plane with the hypotenuse along the line $u=v$, from $(0,0)$ to $(y,y)$ and base along the $v$ axis.  To get the inequality, the integral is over the sector of radius $y$ within the triangle.
