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.

General result:  Let $G(y)=P(|X|\le y)=\frac{2}{\sqrt{2\pi}}\int_0^ye^{-\frac{u^2}{2}}du$. $G^2(y)=\frac{2}{\pi}\int_0^y\int_0^ye^{-\frac{u^2+v^2}{2}}dudv$.
Switch to polar coordinates (note) and get $G^2(y)\gt \int_0^ye^{-\frac{r^2}{2}}rdr=1-e^{-\frac{y^2}{2}}$ or $G(y)\gt (1-e^{-\frac{y^2}{2}})^\frac{1}{2}$.
In general $P(max|X_1|,|X_2|,....,|X_M|\le y)=G^M(y)\gt (1-e^{-\frac{y^2}{2}})^\frac{M}{2}$.  Therefore $P(max|X_1|,|X_2|,....,|X_M|\gt y)\lt 1-(1-e^{-\frac{y^2}{2}})^\frac{M}{2}=1-1+\frac{M}{2}e^{-\frac{y^2}{2}}-...\lt \frac{M}{2}e^{-\frac{y^2}{2}}$
Note:  The domain of integration is a square $y$ by $y$.  The switch to polar coordinates is an integration over the maximum sector within the square, radius $=y$ and angle $=\frac{\pi}{2}$.
