Problem: For $a,b,c\ge0: ab+bc+ca>0.$ Prove that: $$2(a+b+c)+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\ge\sqrt{5ab+4ac}+\sqrt{5bc+4ba}+\sqrt{5ca+4cb}$$
Recently, i have seen a post on AoPS link
My approach: After squaring both side, i get: $$2(a^2+b^2+c^2)+\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})+2(a+b+c)(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\ge\sum{\sqrt{(5ab+4ac)(5bc+4ba)}}$$
Since: $a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}$. It is desired to prove: $$2(a^2+b^2+c^2)+5\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})+2(ab+bc+ca)\ge\sum{\sqrt{(5ab+4ac)(5bc+4ba)}}$$
But it is more complicated. I hope we can find a good approach for problem. Thanks!
Best Answer
Proof.
By denoting $a=x^2;b=y^2;c=z^2$, we will prove $$2\sum_{cyc} x^2 + \sum_{cyc} yz \ge \sum_{cyc} x\sqrt{5z^2+4y^2}.$$ By using Cauchy-Schwarz $$\sum_{cyc} x\sqrt{5z^2+4y^2}=\sum_{cyc} \sqrt{x}\sqrt{5xz^2+4xy^2}\le \sqrt{\sum_{cyc} x .\sum_{cyc} (5xz^2+4xy^2)}.$$ Hence, it suffices to prove $$\left(2\sum_{cyc} x^2 + \sum_{cyc} yz \right)^2\ge \sum_{cyc} x .\sum_{cyc} (5xz^2+4xy^2),$$ which is equivalent to $$4\sum x^4 - \sum x^3y - 3\sum x^2yz \ge 0 .$$The last inequality is true by AM-GM.
Indeed, $$x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2\ge xyz(x+y+z),$$ and $$3.x^4+y^4\ge 4x^3y;3.y^4+z^4\ge 4y^3z;3.z^4+x^4\ge 4z^3x.$$ Hence, the result follows. Equality holds at $a=b=c.$