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## Weird Inequality

Prove that given $a,b,c>0$ we have
$\sqrt[4]{a+\sqrt[3]{b+\sqrt{c}}} > \sqrt[40]{abc}$.

Solution:
It’s easy to see that the inequality is equivalent to $a+\sqrt[3]{b+\sqrt{c}} > \sqrt[10]{abc}$. ( the original form is just for the impression 🙂 )
It may seem difficult to work with these nasty radicals, but there is a trick that blows up the whole thing… 🙂
Let’s see what happens if we replace $(a,b,c)$ with $(ka,k^3b,k^6c)$ where $k>0$ ? We get both sides multiplied with $k$, which shows that if we prove the inequality for one of the triplets $(ka,k^3b,k^6c)$ then the inequality is true for $(a,b,c)$ also. Picking $k=\frac{1}{\sqrt[6]{c}}$, we get $c=1$, so if we prove the inequality for $c=1$ we are done.

Therefore, we have to prove that $a+\sqrt[3]{b+1} - \sqrt[10]{ab} > 0$, for all $a,b >0$. This is an easy calculus exercise, because, fixing $b>0$ we get a function which has a strictly decreasing derivative, so it has a unique minimum point, which is positive.