# 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 $(a,b,c)=(1,1,0)$.

• I assume the condition should be $ab+ac+bc \ne 0$? – Martin R Dec 22 '16 at 18:41
• @Martin R Thank you! I fixed my post. – Michael Rozenberg Dec 22 '16 at 18:56
• It's homogeneous, you can suppose $a=1$ or $ab+ac+bc=73$ for example – Gribouillis Dec 22 '16 at 19:04

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.

• Thank you very much! I thought always that this theorem is not useful. – Michael Rozenberg Jun 28 at 15:16
• Agree. The conditions are strict. I only used it two times. – River Li Jun 28 at 15:20