If $x \sim \mathcal{N}(0, I)$ is an $n$ dimensional random vector. I am looking for a proof of $$ \mathrm{Prob}\left( \lVert x\rVert^2 \ge 6n \right) \le e^{-1.5n}$$
$\lVert x \rVert^2 \sim \chi^{2}$ with $n$ degrees of freedom. So we need a tail bound on a Chi squared random variable.
Using Chebyshev's inequality \begin{align} \mathbb{P}(\|X\|^2 \geq k) \leq e^{-tk}\mathbb{E}[e^{t\|X\|^2}] \end{align} Since $\|X\|^2$ is chi-squared distributed with $n$ degrees of freedom, $$ \mathbb{E}[e^{t\|X\|^2}] = (1-2t)^{-\frac{n}{2}}, \quad \forall t < \frac{1}{2}$$ Substituting $k = 6n $ in the first inequality, \begin{align} \mathbb{P}(\|X\|^2 \geq 6n) &\leq e^{-6tn}(1-2t)^{-\frac{n}{2}}\\ &= \exp\left[-n\left(6t+ \frac{1}{2}\log(1-2t)\right)\right]\\ &\leq \exp\left[- \frac{5 - \log 6}{2} n \right] \textrm{ setting } t = \frac{5}{12}\\ &\leq e^{- 1.5n}\\ \end{align}