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

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

  • $\begingroup$ Thank you very much! I thought always that this theorem is not useful. $\endgroup$ – Michael Rozenberg Jun 28 at 15:16
  • 1
    $\begingroup$ Agree. The conditions are strict. I only used it two times. $\endgroup$ – River Li Jun 28 at 15:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.