Show that $e^x \geq (3/2) x^2$ for all non-negative $x$ 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? 

Use AM-GM (i. e. $a+b \ge 2\sqrt{ab})$:
$$ x+x^3/6 \ge 2\sqrt{x^4/6}=x^2\sqrt{2/3}$$
$$1+x^4/24 \ge 2\sqrt{x^4/24}=x^2\sqrt{1/6}$$
Thus for all $x \ge 0$:
$$e^x \ge 1+x+x^2/2+x^3/6+x^4/24\ge x^2(1/2+ \sqrt{2/3}+\sqrt{1/6}) \ge 3/2 x^2$$
