For all $a,b,c\ge 0$ prove that$$\color{black}{5(a^2+b^2+c^2)+ab+bc+ca\ge 2\sum_{cyc}a\sqrt{5a^2+b^2+c^2+ab+ac}.}$$
I've tried to use AM-GM without success.
Indeed, $$3\cdot RHS=2\sum_{cyc}3a\sqrt{5a^2+b^2+c^2+ab+ac}\le \sum_{cyc}\left(9a^2+5a^2+b^2+c^2+ab+ac\right)=16(a^2+b^2+c^2)+2(ab+bc+ca) $$
It remains to prove $$15(a^2+b^2+c^2)+3(ab+bc+ca)\ge16(a^2+b^2+c^2)+2(ab+bc+ca) $$
$$\iff a^2+b^2+c^2\le ab+bc+ca.$$
It is obviously wrong since the last inequality is equivalent to $$(a-b)^2+(b-c)^2+(c-a)^2\le 0.$$
Could you please help me prove the starting inequality? Thank you for your help.
Best Answer
Proof.
By using AM-GM as \begin{align*} 2\sqrt{5a^2+b^2+c^2+ab+ac}&=2\sqrt{\frac{5a^2+b^2+c^2+ab+ac}{\sqrt{bc}+2a}\cdot\left(\sqrt{bc}+2a\right)}\\&\le \frac{5a^2+b^2+c^2+ab+ac}{\sqrt{bc}+2a}+\sqrt{bc}+2a\\&=\frac{bc+a^2+ab+ac+b^2+c^2}{\sqrt{bc}+2a}+4a. \end{align*}Also,$$2a+\sqrt{bc}\ge 2a+\frac{2bc}{b+c}=\frac{2(ab+bc+ca)}{b+c}.$$ Subsequently, $$\frac{(ab+ac)[bc+ab+ac+a^2+b^2+c^2]}{2(ab+bc+ca)}+4a^2\ge 2a\sqrt{5a^2+b^2+c^2+ab+ac}.$$ Sum up analogs, we get desired inequality. Equality holds iff $a=b=c.$