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

Question

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.

Answer 1

Firstly $b+\sqrt{c}=b+\sqrt{c}/2+\sqrt{c}/2\ge 3\sqrt[3]{bc/4}$.

Then $a + \sqrt{b + \sqrt{c}}\ge a+\sqrt{3}\sqrt[6]{bc/4}=a+\frac{\sqrt{3}\sqrt[6]{bc/4}}{6}+\cdots+\frac{\sqrt{3}\sqrt[6]{bc/4}}{6}\ge 7\sqrt[7]{Cabc}$.

Here $C$ is a constant which equals $\frac{1}{4}(\sqrt{3}/6)^6$. Then just need to check $7\sqrt[7]{C}\ge 1$ to see why $n=14$ works. And you can show that this is the minimized $a + \sqrt{b + \sqrt{c}}$ when you fix $abc$, which proves other $n$ fails.

Sorry for the messy notion, hope you can grasp the idea.

Answer 2

$\textbf{Hint:}$ Use a change of variables such that

$$\begin{cases}u=a\\ v^2=b+\sqrt{c}\\ w=b-\sqrt{c}\end{cases}\implies u+v\geq\left(\frac{u(v^4-w^2)(v^2-w)}{8}\right)^{\frac{2}{n}}$$

This means we can reduce this to the problem, finding which $n$ does the following inequality hold?

$$u+v\geq\left(\frac{u(v^4-w^2)(v^2-w)}{8}\right)^{\frac{2}{n}}$$

Answer 3

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 $n = 14$?

Answer 4

After beautiful Peter Wu's and Jim Ferry's solutions,

I want to add a bit of simpler proof for $n=14.$

Let $\sqrt{a+\sqrt{b+\sqrt{c}}}=x.$

Thus, $$x\geq\sqrt{a},$$ $$x\geq\sqrt[4]{b}$$ and $$x\geq\sqrt[8]c.$$ Thus, $$x^{2+4+8}\geq abc$$ and $$\sqrt{a+\sqrt{b+\sqrt{c}}}\geq\sqrt[14]{abc}.$$