Prove $\frac{a}{\sqrt{b^2+c^2+7bc}}+\frac{b}{\sqrt{c^2+a^2+7ca}}+\frac{c}{\sqrt{a^2+b^2+7ab}} \le \frac{a^2+b^2+c^2}{ab+bc+ca} a,b,c \ge 0 ab+bc+ca>0$

Question

Let $a,b,c \ge 0 : ab+bc+ca>0.$ Prove that $$\frac{a}{\sqrt{b^2+c^2+7bc}}+\frac{b}{\sqrt{c^2+a^2+7ca}}+\frac{c}{\sqrt{a^2+b^2+7ab}} \le \frac{a^2+b^2+c^2}{ab+bc+ca}.$$

I posted on AOPS here.

I am the author of this problem. I hope someone prove it by a nice solution. I saw this proof by Cauchy - Schwarz: $$\sum\limits_{cyc}\frac{a}{\sqrt{b^2+c^2+7bc}} \le \sqrt{(a+b+c)\sum\limits_{cyc}\frac{a}{b^2+c^2+7bc}}.$$ It suffices to prove that $$\sum\limits_{cyc}\dfrac{a}{b^2+c^2+7bc} \le \dfrac{(a^2+b^2+c^2)^2}{(ab+bc+ca)^2(a+b+c)}.$$ I think this is true, but I can't see any nice solution to prove it. Please help me.

Answer 1

Your way leads to a wrong inequality.

Try $(a,b,c)=(3,1,0).$

My solution:

Since by C-S $$\sqrt{b^2+7bc+c^2}=\frac{\sqrt{\left((b-c)^2+4bc\right)\left((b-c)^2+9bc\right)}}{b+c}\geq$$ $$\geq\frac{(b-c)^2+6bc}{b+c}=\frac{b^2+4bc+c^2}{b+c},$$ it's enough to prove that: $$\sum_{cyc}\frac{a(b+c)}{b^2+4bc+c^2}\leq\frac{a^2+b^2+c^2}{ab+ac+bc}$$ or $$abc\sum_{sym}(a^5+7a^4b-4a^3b^2+8a^3bc-12a^2b^2c)\geq0,$$ which is true by Muirhead.