Prove that $e^{-x^2}<\frac{1}{x^3}$, if $x>0$

Question

I have to prove that if $x>0$ then $e^{-x^2}<\frac{1}{x^3}$.
I have thought to set $g(x):=e^{-x^2}-\frac{1}{x^3}$ and from this $g'(x)=-2xe^{-x^2}+\frac{3}{x^4}$, but I think this idea is not good since now I can't prove easily that $g'(x)<0$. Can you help me?

Answer 1

Use that $e^x \geq 1 + x$ for all $x \in \mathbb R$. Then $$e^{x^2/3} \geq 1 + \frac{x^2}3 > x$$ for all $x\in \mathbb R$. Now raise both sides to the third power to get $$e^{x^2} > x^3$$ for all $x \in \mathbb R$.

Answer 2

Edit: Now that the question has been updated, I will remove my (no longer relevant) counterexample.

It is enough to prove for $x>0$, that $x^3<e^{x^2}$, which you can prove by differentiating $g(x)=e^{x^2}-x^3$.

$g'(x)=2xe^{x^2}-3x^2$

$g''(x)=(2+4x^2)e^{x^2}-6x$

$g'''(x)=(12x+8x^3)e^{x^2}-6$

$g''''(x)=(12+48x^2+16x^4)e^{x^2}$

Now $g^{(4)}$ is strictly positive for $x>0$, and $g^{(3)}(0)=0$, so $g^{(3)}$ is strictly positive for $x>0$. Similarly, $g''(0)=g'(0)=g(0)=0$, so we can continue this reasoning to obtain that $g$ is strictly positive for $x>0$, which yields the desired inequality.

Answer 3

For $x > 0$ is, using the Taylor series for the exponential function and the inequality between arithmetic and geometric mean: $$ \begin{align} e^{x^2} &> 1 + x^2 + \frac {1}{2!} x^4 + \frac {1}{3!} x^6 \\ &= \frac 14 \left( 4 + 4 x^2 + 2 x^4 + \frac 23 x^6\right) \\ &\ge \sqrt[4]{4 \cdot 4 x^2 \cdot 2 x^4 \cdot \frac 23 x^6} \\ & > \sqrt[4]{16 x^{12}} = 2 x^3 \, , \end{align} $$ which gives the slightly better estimate $$ e^{-x^2} < \frac{1}{2x^3} \, . $$

Answer 4

Solution 1 Let $h(x)=x^3e^{-x^2}.$ Then $$h'(x)=3x^2e^{-x^2}-2x^4e^{-x^2}=x^2e^{-x^2}[3-2x^2]$$ The maximal value for $x>0$ is attained at $x=(3/2)^{1/2}.$ Hence $$x^3e^{-x^2}\le a:= {\sqrt{3}\over \sqrt{2}}\,{3\over 2}e^{-3/2}$$ That's a best possible constant. The function $xe^{-x}$ attains the maximum at $x=1$ and is equal $e^{-1}. $ Hence $${\sqrt{2}\over \sqrt{3}}\,{3\over 2}e^{-3/2}\le {\sqrt{3}\over \sqrt{2}}e^{-1}={\sqrt{6}\over 2e}<{1\over 2}$$ According to WolphramAlpha ${\sqrt{6}\over 2e}<0.451.$ On the other hand $0.4<a<0.41.$ So the estimate given by @Martin R is pretty close to the optimal one.

Solution 2 For $0\le x< 1$ we have $x^3e^{-x^2}<1.$ For $x\ge 1$ there holds $x^3e^{-x^2}\le x^4e^{-x^2}.$ So it suffices to estimate $t^2e^{-t}$ for $t\ge 1.$ The maximal value is attained at $t=2$ and is equal $4e^{-2}<1.$

Answer 5

Let $f(x)=x^3e^{-x^2}$. We want to show that $f(x)<1$ for all $x>0$.

$f$ is maximized when $f'(x)=0$.

$$f'(x)=3x^2e^{-x^2}+x^3(-2xe^{-x^2})=x^2e^{-x^2}(3-2x^2)$$

and this is $0$ when $x=\sqrt{3/2}$.

At this max we have $f(\sqrt{3/2})=(3/2)^{3/2}e^{-3/2}=(1.5/e)^{3/2}$. This is obviously less than 1 since $e>1.5$.

Since $f<1$ at its maximum value, it must be so for all $x>0$.