Prove that ${\sqrt {n} }^{\sqrt {n+1}} > {\sqrt {n+1}}^{\sqrt {n}}$.

Which is greater? $\sqrt{n}^{\sqrt{n+1}}$ or $\sqrt{n+1}^\sqrt{n}$

I know that $\sqrt{n}^{\sqrt{n+1}}$ is greater but I tried using induction and I couldn't figure it out. Thanks for the help.

Solution 1:

Note that $\frac{\mathrm{d}}{\mathrm{d}x}\frac1x\log(x)=\frac1{x^2}(1-\log(x))<0$ for $x\gt e$. Therefore, $x^{1/x}$ is a decreasing function for $x\gt e$. Thus, for $\sqrt{n}\ge e$, i.e. $n\gt e^2$, $$ \sqrt{n+1}^{1/\sqrt{n+1}}\lt\sqrt{n}^{1/\sqrt{n}}\\ \Updownarrow\\ \sqrt{n+1}^{\sqrt{n}}\lt\sqrt{n}^{\sqrt{n+1}} $$

Solution 2:

It actually depends on what $n$ is.

Consider a fixed $a$, and let $f(x)=x^a$ and $g(x)=a^x$. At $x=a$, both functions have equal output. Let's examine their derivatives at $x=a$: $$f'(a) = a^a$$ $$g'(a) = a^a\ln(a)$$ Generally $g$ has the larger derivative at $x=a$, as long as $a>e$. Again as long as $a>e$, after the point at which $f$ and $g$ are equal, $g$ has higher output values.

So let's take $a=\sqrt{n}$ and assume for a bit that $\sqrt{n}>e$. Since $\sqrt{n+1}$ is greater than $a$, we will have that $g(\sqrt{n+1})>f(\sqrt{n+1})$, or rather: $$\sqrt{n}^{\sqrt{n+1}}>\sqrt{n+1}^{\sqrt{n}}$$

For values of $n$ smaller than $e^2$, you could just examine those $n$ values individually. For $n=7$, this direction is still true, but the inequality reverses for $n=0\ldots6$.

Solution 3:

This is just another version of "$n^{1/n}$ is decreasing for $n > e$" that pops up here every so often. I and others have provided a number of proofs of this.