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? 

Consider a quadratic $y=a x^2$ that is tangent to $y=e^x$. That is, we match function and derivative at a point $x=x_0$.  Then
$$e^{x_0}=a x_0^2$$
$$e^{x_0}=2 a x_0$$
Then $a x_0^2=2 a x_0$ so that $x_0=2$.  Plugging this back into one of the above equations, we get that
$$a=\frac{e^2}{4} \approx 1.84726$$
Consider, as @julien suggests, the function $g(x)=a x^2 e^{-x}$ which has derivative $a (2-x) x e^{-x}$ and second derivative $a (x^2-4 x+2) e^{-x}$.  This function represents the ratio of the quadratic to the exponential and is a maximum at $x=2$ for all $a>0$.  As $a x^2$ monotonically increases in $a$, we have
$$e^x \ge \frac{e^2}{4} x^2 \gt \frac{3}{2} x^2$$
