If $Z$ is a centered normal random variable with variance $\sigma^2$ (hence sub-gaussian also), how can I show that
$$ \sup\limits_{t>0} \Bigg( P(Z \geq t)\exp{\Big(\frac{t^2}{2\sigma^2}}\Big) \Bigg) = \frac{1}{2} $$ I tried using the fact that $P(Z \geq t) = \int\limits_{t}^{\infty} f(z)dz$, but it did not help to find a closed form.
Hint. The conditional expectation $E[Z|Z\ge t]$ is obviously $\ge t$ and is easily calculable by calculus.
In particular, let $\sigma=$ without loss of generality, and write $P(Z\in A)=\int_A \varphi(x)\,dx $ where $\varphi(x)=\exp(-x^2/2)/\sqrt{2\pi}$. Let $h(t)=P(Z\ge t)\exp(t^2/2)$ so $$h'(t)=-\varphi(t)\exp(t^2/2)+P(Z\ge t)t\exp(t^2/2).$$ Since $h(0)=1/2$, it suffices to show $h'(t)\le 0$ for $t\le0$, that is, $$P(Z\ge t)t\exp(t^2/2)\le \varphi(t)\exp(t^2/2),$$ that is $$t\le\frac{\varphi(t)}{P(Z\ge t)}\tag{*}.$$ But $$E[Z|Z\ge t]P(Z\ge t) = \int_t^\infty x \varphi(x)\,dx = \frac1{\sqrt{2\pi}}\int_t^\infty x e^{-x^2/2}\,dx = \varphi(x),$$ so (*) becomes $t\le E[Z|Z\ge t],$ which clearly holds.