Very hard AM – GM inequality question: $\sqrt{a + \sqrt{b + \sqrt{c}}} \ge \sqrt[n]{abc}$

a.m.-g.m.-inequalityalgebra-precalculuscontest-mathinequality

Find the sum of all positive integers $n,$ where the inequality
$$\sqrt{a + \sqrt{b + \sqrt{c}}} \ge \sqrt[n]{abc}$$holds for all nonnegative real numbers $a,$ $b,$ and $c.$

What I have tried:

I have already tried squaring both sides but that got me nowhere.

I also have tried the AM – GM inequality and this is what I have:

$$\frac {a + b + c + (n – 3) \cdot 1} {n} \ge \sqrt [n] {abc}.$$

I'm mainly looking for hints, but answers are welcome.

Best Answer

The first thing I would do is introduce variables that I can think of as "being in the same units" (e.g., meters). Thus, let $a = u$, $b = v^2$, and $c = w^4$. Then we may write the inequality as$$ u + \sqrt{v^2 + w^2} \ge x^{14/n}, \;\text{where}\; x = (u v^2 w^4)^{1/7}. $$ Here we are thinking of $u$, $v$, $w$, and $x$ as being quantities in meters. If $n = 14$ then both sides of the inequality are in the same units. But otherwise the two sides are in different units, which seems strange. How can we exploit this? E.g., if $n < 14$ what happens when $u$, $v$, and $w$ are very large? What happens when $n > 14$? Finally, what happens when $n = 14$?