Prove that $\sum\limits_{cyc}\frac{1}{\sqrt{a^2+ab+b^2}}\geq\frac{2}{\sqrt{ab+ac+bc}}+\sqrt{\frac{a+b+c}{3(a^3+b^3+c^3)}}$ Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
$$\frac{1}{\sqrt{a^2+ab+b^2}}+\frac{1}{\sqrt{a^2+ac+c^2}}+\frac{1}{\sqrt{b^2+bc+c^2}}\geq\frac{2}{\sqrt{ab+ac+bc}}+\sqrt{\frac{a+b+c}{3(a^3+b^3+c^3)}}$$
I tried C-S, Holder and more, but without success.  
The equality occurs here also for $(a,b,c)=(1,1,0)$. 

When one of $a, b, c$ is zero, clearly the inequality is true. In the following, assume that $a, b, c > 0$.
We apply Ji Chen's Symmetric Function Theorem for $n=3$:
(see https://artofproblemsolving.com/community/c6h194103p1065812)
Let $d\in (0,1)$. Let $x, y, z, u, v, w$ be non-negative real numbers satisfying
$$x+y+z \ge u+v+w, \quad
xy+yz+zx \ge uv+vw+wu, \quad
xyz \ge uvw.$$
Then $x^d + y^d+z^d \ge u^d + v^d+w^d$.
Now let us prove the inequality. Let 
\begin{align}
&X = \frac{1}{a^2+ab+b^2}, \ Y = \frac{1}{b^2+bc+c^2}, \ Z = \frac{1}{c^2+ca+a^2},\\
&U = V = \frac{1}{ab+bc+ca}, \ W = \frac{a+b+c}{3(a^3+b^3+c^3)}.
\end{align}
We need to prove that $\sqrt{X} + \sqrt{Y} + \sqrt{Z} \ge \sqrt{U} + \sqrt{V} + \sqrt{W}$.
Let 
\begin{align}
f = X + Y + Z - (U+V+W), \ g = XY+YZ+ZX - (UV+VW+WU), \ h = XYZ - UVW.
\end{align}
We can prove that $f, g, h\ge 0$ using Buffalo Way. 
To prove that $f \ge 0$, it suffices to prove that $f_1(a,b,c) \ge 0$ where $f_1(a,b,c)$ is a polynomial.
WLOG, assume that $a\ge b\ge c = 1$. Note that $f_1(1+s+t, 1+s, 1)$ is a polynomial in $s, t$ with non-negative coefficients. The inequality is true.
Similarly, we may prove that $g\ge 0$ and $h\ge 0$.
According to Ji Chen's Symmetric Function Theorem, we are done.
