show this inequality $\sum_{cyc}\sqrt{a(b+c)}(b^2+c^2-a^2-bc)\ge 0$

Question

let $a,b,c>0$,show this inequality $$\sqrt{a(b+c)}(b^2+c^2-a^2-bc)+\sqrt{b(c+a)}(c^2+a^2-b^2-ca)+\sqrt{c(a+b)}(a^2+b^2-c^2-ab)\ge 0$$

I want use S-O-S methods to solve ,But I can't, see this methods:https://artofproblemsolving.com/community/c6h80127

so My inequality How to do?Thanks

Answer 1

SOS method works:

Remarks: I update my proof. My proof works for $a, b, c > 0$.

We need to prove that $$\sum_{\mathrm{cyc}} 2\sqrt{a(b + c)}(b^2 + c^2 - a^2 - bc) \ge 0$$ or $$\sum_{\mathrm{cyc}} \sqrt{a(b + c)}\,\left[2c(c - a) - 2b(a - b) - (a - b)^2 + (b - c)^2 - (c - a)^2\right] \ge 0$$ or \begin{align*} &\sum_{\mathrm{cyc}} \left[\left(2a\sqrt{b(c + a)} - 2b\sqrt{a(b + c)}\right) (a - b) \right. \\ &\quad + \left. \left(-\sqrt{a(b + c)}- \sqrt{b(c + a)} + \sqrt{c(a + b)}\right)(a - b)^2\right]\ge 0 \end{align*} or $$\sum_{\mathrm{cyc}} \left[-(a^2 - b^2)\sqrt{a(b + c)} + (a^2 - b^2)\sqrt{b(c + a)} + (a - b)^2\sqrt{c(a + b)}\right] \ge 0$$ or $$\sum_{\mathrm{cyc}} \left[-(a^2 - b^2)\frac{(a - b)c}{\sqrt{a(b + c)} + \sqrt{b(c + a)}} + (a - b)^2\sqrt{c(a + b)}\right] \ge 0$$ or $$\sum_{\mathrm{cyc}} \left[1 - \frac{\sqrt{c(a + b)}}{\sqrt{a(b + c)} + \sqrt{b(c + a)}}\right]\sqrt{c(a + b)}\,(a - b)^2 \ge 0$$ or $$\sum_{\mathrm{cyc}} \frac{2ab\sqrt{c(a +b)} + 2\sqrt{abc(a + b)(b + c)(c + a)}}{[\sqrt{a(b + c)} + \sqrt{b(c + a)}]\,[\sqrt{a(b + c)} + \sqrt{b(c + a)} + \sqrt{c(a + b)}]}\,(a - b)^2 \ge 0$$ which is clearly true.

We are done.