TITLE: Show that $e^x \geq (3/2) x^2$ for all non-negative $x$
QUESTION [11 upvotes]: I am attempting to solve a two-part problem, posed in Buck's Advanced Calculus on page 153. It asks "Show that $e^x \geq \frac{3}{2}x^2$  $\forall x\geq 0$. Can $3/2$ be replaced by a larger constant?"
This is after the section regarding Taylor polynomials, so I have been attempting to leverage the Taylor expansion for $e^x$ at $0$. $e^x \geq 1+x+\frac{x^2}{2}$, and by quadratic formula, we have $1+x+\frac{x^2}{2}\geq \frac{3}{2}x^2$ for $x\in [0, \frac{1+\sqrt{5}}{2}]$.
Now also $e^x\geq 1+x+\frac{x^2}{2}+\frac{x^3}{6}$. We know there exists a $c\in \mathbb{R}^{\geq 0}$ such that for all $x\geq c$, we have $1+x+\frac{x^2}{2}+\frac{x^3}{6}\geq \frac{3}{2}x^2$. I want to find this point without messing with the cubic formula, etc. I think I am missing a simpler way. Any ideas?

REPLY [10 votes]: Start from $e^x \geq 1 + x$ for all $x \in \mathbb{R}$. Then $$e^x = e \cdot e^{x-1} \geq e \cdot x$$ for all $x \in \mathbb{R}$ and  $$e^x = (e^{\tfrac{x}{2}})^2 \geq \left(\frac{e \cdot x}{2} \right)^2$$ for all $x \geq 0$.