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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}.

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.

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