Proving that $e^x \geq \sqrt{x}$ for all $x\geq0$

Question

Prove that $e^x \geq \sqrt{x}$ for all $x\geq 0$.

My attempt: since $e^x>0$ for all $x\in\mathbb{R}$ and $\sqrt{x} \geq 0$ for all $x\geq0$, the inequality is equivalent if we square both sides; so the initial statement is equivalent to prove that $e^{2x} \geq x$.

We know from the Taylor expansion that $e^{2x} \geq 1+2x$ for all $x\in\mathbb{R}$ and it is $1+2x\geq2x$; since for hypothesis $x\geq0$, we have that $2x \geq x$.

So we end up with $e^{2x} \geq 1+2x \geq x$ and this is what we wanted to prove.

Is this right? If it isn't where is the error?

If it is right, any other ways to prove it are welcome. Thanks to you all.

Answer 1

Your proof looks correct. Here is another way to prove this without using advanced result from Taylor expansion.

For $x=0$, you have $e^{0}=1>0=\sqrt{0}$.

For $x>0$, note that $e^{x}>0$ and $\sqrt{x}>0$ so we can take the natural $\log$, so that $$\log(e^{x})=x\ \text{and}\ \log(\sqrt{x})=\frac{1}{2}\log(x).$$

Then set $g(x):=x-\dfrac{1}{2}\log x$. Take derivate you will find $g'(x)>0$ for all $x>1$.

Thus, $x>\dfrac{1}{2}\log x$ for all $x>1$.

At $x=1$, we have $\sqrt{1}=1>\log(1)=0$.

At $x\in (0,1)$, we have $x>0$ but $\log(x)<0$.

Thus, we can conclude that $\log(e^{x})>\log(\sqrt{x})$ for all $x\geq 0$. Since, $\log$ is an increasing (non-decreasing) function in $(0,\infty)$, we can conclude that $e^{x}\geq \sqrt{x}$ for all $x\geq 0$.

Answer 2

It is completely right.

Another way could be:

• $x\in [0,1] \Rightarrow e^x\geq 1 \geq \sqrt x$
• $x \in [1,+\infty) \Rightarrow e \geq 1$ and $\left(e^x-\sqrt x\right)' = e^x - \frac 1{2\sqrt x}\geq e-\frac 12 >0$.

Answer 3

Also, we have $$e^x\geq1+x>\sqrt{x}.$$