# Prove that $\frac{x^2}{4} \leq e^{x - 2}$ for all $x \geq 0$

I don't have too many tools for proving inequalities. It's an easy application of the racetrack principle to show this result for $x \geq 2$. I am struggling with showing the result for $0 \leq x \leq 2$ We see that $\frac{x^2}{4}$ is tangent to $e^{x-2}$ at $x = 2$. Really the only tools I have for proving inequalities is comparison, induction and the racetrack principle. I might be able to use the taylor series for $e^{x - 2}$ but it seems fairly complicated since for $x \in [0,2)$, the terms alternate. Any insight or hints are greatly appreciated.

Use $e^x\geq x+1$ for all $x\in\mathbb{R}$. Substitute $x\leftarrow x/2-1$ to get $e^{x/2-1}\geq x/2$ and square both sides (here use that $x\geq 0$).
Taking the $\log$ of both sides, we want to show $$x - 2 \geq 2 \log \frac x2$$ or
$$u - 1 \geq \log u$$
Since $\log$ is concave, it is bounded above by its first order taylor approximation at $u=1$.