A quite delicious inequality problem

[anemone]

Prove that $2^{2\sqrt{3}}>10$.

 

[kaliprasad]

Spoiler: My solution

we have $3 > \frac{25}{9}$
or $\sqrt{3} > \frac{5}{3}$

hence $2^{2\sqrt{3}} > 2^{2*\frac{5}{3}}\cdots(1)$

Now $2^{2*\frac{5}{3}} =2^{\frac{10}{3}}= \sqrt[3]{2^{10}}= \sqrt[3]{1024}> 10\cdots(2) $

from (1) and (2) we get above