$\left[\begin{array}{l}\text { Numerical inequalities. Comparing numbers. } \\ \text { [ Classical inequalities (miscellaneous). ] }\end{array}\right.$

Find the largest of the numbers $5^{100}, 6^{91}, 7^{90}, 8^{85}$.
$\left({ }^{6} / 5\right)^{9}=(216 / 215)^{3}>\left({ }^{216} / 126\right)^{3}=\left({ }^{12 / 7}\right)^{3}=1728 / 343>5$. Therefore, $6^{90}>5^{100}$.

According to Bernoulli's inequality $(7 / 6)^{90}>1+{ }^{90} / 6=16>6$. Therefore, $7^{90}>6^{91}$.

Notice that ${ }^{m / n}>{ }^{m+1 / n+1}$ when $m>n>0$. Therefore, $\left({ }^{8} / 7\right)^{6}=\left({ }^{64} / 49\right)^{3}>\left({ }^{65} / 50\right)^{3}=(1.3)^{3}=2.197>2$.

Therefore, $8^{6}>2 \cdot 7^{6}, 8^{90}>2^{15} \cdot 7^{90}=8^{5} \cdot 7^{90}$, which means $8^{85}>7^{90}$.

## Answer

$8^{85}$