<s>[INST] 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.
 [/INST] 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}}$.
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