Have You Seen This Inequality Before?

With $x>0$ , choose $a=x^2; b=x^4; c=x^8$ , we get: $x\sqrt{1+\sqrt{2}}\ge x^{\frac{14}{n}}\qquad(*)$ If $n<14$ , $(*)\Leftrightarrow \sqrt{1+\sqrt{2}}\ge x^{\frac{14-n}{n}}$ , this inequality is false where $x$ is enough large.

If $n>14$ , $(*)\Leftrightarrow x^{\frac{n-14}{n}} \sqrt{1+\sqrt{2}}\ge 1$ , this inequality is false where $x$ is enough small.

We will prove that the initial inequality is true where $n=14$ .

Indeed,

$\quad\left(\sqrt{a+\sqrt{b+\sqrt c}}\right)^{14}$

$\displaystyle=\left(a+\sqrt{b+\sqrt c}\right)^7=\sum_{i=0}^7 \binom{7}{i} a^i\left(\sqrt{b+\sqrt c}\right)^{7-i}\ge 7a\left(\sqrt{b+\sqrt c}\right)^6$

$\displaystyle=7a\left(b+\sqrt c\right)^3=7a\sum_{i=0}^3 \binom{3}{i} b^i\left(\sqrt c\right)^{3-i}\ge 21abc\ge abc$

So, $\boxed{n=14}$ is only satisfied value.