$\frac{1}{3}\sum_{cyc}\frac{1}{\sqrt{1+a}}\ge\frac{1}{\sqrt{1+\sqrt[3]{abc}}}$ if $\;a, b, c\;$ are positive reals s.t. $abc\ge2^9$

Question

$$\frac{1}{3}\sum_{cyc}\frac{1}{\sqrt{1+a}}\ge\frac{1}{\sqrt{1+\sqrt[3]{abc}}}$$ if it is given that $\;a, b, c\;$ are positive reals s.t. $abc\ge2^9$.

I have tried (many) dead-end solutions. Initially setting $\;abc\;$ to be greater than $512$ on the left-hand-side to make it $\frac{1}{3}$ lead to an untrue inequality.

Letting $f(x)=\frac{1}{\sqrt{1+x}}$ we find that $LHS=\frac{1}{3}\big(f(a)+f(b)+f(c)\big)$ and, since $f''\ge0$, $f$ is convex and, by Jensen's inequality, $$\frac{1}{3}\big(f(a)+f(b)+f(c)\big)\ge f\left(\frac{a+b+c}{3}\right)= \frac{1}{\sqrt{1+\frac{a+b+c}{3}}},$$ but the result is no longer greater than the original RHS; it is in fact less than or equal to the original RHS, as can be proven with a simple application of AM-GM.

Do you have any hints on what to try next? I feel that I have exhausted all of my ideas at this point.
Thanks beforehand!

Answer 1

Let $a=\frac{8kyz}{x^2},$ $b=\frac{8kxz}{y^2}$ and $c=\frac{8kxy}{z^2},$ where $x$, $y$, $z$ and $k$ are positive numbers.

Thus, the condition gives $k\geq1$ and we need to prove that $$\sum_{cyc}\frac{x}{\sqrt{x^2+8xyz}}\geq\frac{3}{\sqrt{1+8k}}.$$ Now, by Holder $$\left(\sum_{cyc}\frac{x}{\sqrt{x^2+8xyz}}\right)^2\sum_{cyc}x(x^2+8kyz)\geq(x+y+z)^3.$$ Id est, it's enough to prove that $$(1+8k)(x+y+z)^3\geq9\sum_{cyc}x(x^2+8kyz)$$ or $$\sum_{cyc}(8(k-1)x^3+3(1+8k)(x^2y+x^2z)-(56k-2)xyz)\geq0$$ or $$8(k-1)\sum_{cyc}(x^3-x^2y-x^2z+xyz)+(32k-5)\sum_{cyc}z(x-y)^2\geq0,$$ which is true by Schur.