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

$$\frac{1}{3}\sum_{cyc}\frac{1}{\sqrt{1+a}}\ge\frac{1}{\sqrt{1+\sqrt[3]{abc}}}$$ 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!

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.