calculus - Gaussian Definite Integral for Gaussian random variables

Weh have the following:

For $s>0$, $$\int_s^{\infty} e^{-x^2/2}\,\mathrm{d}x \le \frac 1 s e^{-s^2/ 2}, \quad \int_s^{\infty} e^{-x^2/2}\,\mathrm{d}x \le \sqrt {\frac \pi 2} e^{-s^2/ 2}.$$
Then prove that for a random variable $X \sim N(0,1)$ and $s>0$, $$ \Pr(X>s) \le \frac 1 {\sqrt {2\pi}}\min\left({\frac 1 t, \sqrt {\frac \pi 2}}\right)e^{-s^2/2}.$$

So far i have the following $\Pr(X>s) = \int_s^{\infty} e^{-x^2}\,\mathrm{d}x \le \min\left(\frac 1 t, {\frac \pi 2}\right) e^{-s^2/2},$, which does not need any further proof if i get first the two inequalities right?

I got the first part where it is $\le 1/s$ but for the second part, I tried converting to polar coordinates for the second equation. Since I did $$\int_s^{\infty} e^{-x^2/2}\,\mathrm{d}x \int_s^{\infty} e^{-y^2/2}\,\mathrm{d}y \le \int_0^{\pi/2} \int_s^{\infty} e^{-r^2/2}\,r\mathrm{d}rd\theta \to \int_s^{\infty} e^{-x^2/2}\,\mathrm{d}x \le \sqrt {\frac \pi 2 e^{-s^2/ 2}}$$? Am i doing something wrong?