Can You Prove e^x Is Greater Than x^2 for All Positive x?

[anemone]

Here is this week's POTW:

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Prove \(\displaystyle e^x>x^2\) for all $x>0$.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution!(Cool)

1. kaliprasad
2. Olinguito
3. castor28

Solution from castor28:

Spoiler

If $0<x<1$, $x^2<x$. On the other hand, using the Taylor series for $e^x$, we have:
$$
e^x > 1 + x > x > x^2
$$
and this shows that the relation holds in this case.

We assume now that $x\ge1$ and define $f(x) = e^x-x^2$. We must show that $f(x)>0$. We have:
\begin{align*}
f'(x)&= e^x - 2x\\
f''(x) &= e^x - 2
\end{align*}
As $x\ge1$, $f''(x)\ge e-2>0$.

This shows that $f'(x)$ is an increasing function. As $f'(1) = e-2>0$, $f'(x)>0$ for all $x\ge1$.

This shows that $f(x)$ is increasing; as $f(1)=e-2>0$, $f(x)>0$ for all $x\ge1$, which is what we wanted to prove.