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?

Solution 1:

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 $x \ge 0$:

$$e^x \ge 1+x+x^2/2+x^3/6+x^4/24\ge x^2(1/2+ \sqrt{2/3}+\sqrt{1/6}) \ge 3/2 x^2$$

Solution 2:

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

$$e^x \ge \frac{e^2}{4} x^2 \gt \frac{3}{2} x^2$$

Solution 3:

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 $x \geq 0$.