Prove that $\sqrt{7}^{\sqrt{8}}>\sqrt{8}^{\sqrt{7}}$ show that
$$\sqrt{7}^{\sqrt{8}}>\sqrt{8}^{\sqrt{7}}$$
and  I found
$$LHs-RHS=0.017\cdots$$
I have post this interesting problem Prove $\left(\frac{2}{5}\right)^{\frac{2}{5}}<\ln{2}$
can someone suggest  any other nice method? Thank you everyone.
 A: I just want to give you a bigger picture on this question. The fact you point out corresponds to $f(2)>0.$
$f(\text{x$\_$})\text{:=}\left(t=\lfloor \exp (x)\rfloor ;a=t^{1/x};b=(t+1)^{1/x};a^b-b^a\right);$
$\text{Plot}[f(x),\{x,1,3\}]$

A: Note that $\sqrt{7}^{\sqrt{8}}\doteq15.673$ and $\sqrt{8}^{\sqrt{7}}\doteq15.656$, so these two are pretty close. Furthermore $\sqrt{7}<e<\sqrt{8}$, and the function $f(x):={\log x\over x}$ has a local maximum at $x:=e$. This excludes the use of monotonicity arguments. The following proof uses integer arithmetic instead.
To begin with we need a rational approximation to $\sqrt{7\over8}$ that is slightly larger than $\sqrt{7\over8}$. Using Mathematica (or continued fractions) one finds that $$7\cdot 31^2=6727<6728=8\cdot 29^2\ ,$$ which implies
$$\sqrt{7}\cdot31<29\cdot\sqrt{8}\ .$$
Furthermore one computes
$$8^{29}=15\>47425\>04910\>67253\>43623\>90528<15\>77753\>82034\>84580\>66150\>42743=7^{31}\ .$$
It follows that
$$\bigl(8^{\sqrt{7}}\bigr)^{31}<\bigl(8^{29}\bigr)^{\sqrt{8}}<\bigl(7^{31}\bigr)^{\sqrt{8}}\ .$$
Now take the $62^{\rm th}$ root on both sides.
A: A solution from a greek math forum.
$\frac{lne^{2}-ln7}{e^{2}-7} < \frac{1}{7} \Leftrightarrow \frac{1}{ln7} < \frac{7}{21-e^{2}}<\frac{7}{13,6}$
$\frac{ln8-lne^{2}}{8-e^{2}} >\frac{1}{8} \Leftrightarrow \frac{1}{ln8} < \frac{8}{24-e^{2}}<\frac{8}{16,6}$
So,
$ \frac{1}{ln7}+ \frac{1}{ln8} < \frac{7}{13,6}+ \frac{8}{16,6}= \frac{225}{225,76}<1$.
$\int_{ln7}^{ln8} \frac{1}{x} dx < \frac{(\frac{1}{ln7}+\frac{1}{ln8}) \cdot (ln8-ln7)}{2} < ln \sqrt{8}-ln\sqrt{7}=\int_{\sqrt{7}}^{\sqrt{8}} \frac{1}{x} dx$.
So,
$\int_{ln7}^{ln8} \frac{1}{x} dx < \int_{\sqrt{7}}^{\sqrt{8}} \frac{1}{x} dx\Leftrightarrow $
$ln\frac{ln8}{ln7} < ln \frac{\sqrt{8}}{\sqrt{7}} \Leftrightarrow \sqrt{8}ln7 > \sqrt{7} ln8 \Leftrightarrow$
$ \sqrt{7}^{\sqrt{8}}> \sqrt{8}^{\sqrt{7}}$.
