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!
We need to prove that: $$\sum_{cyc}(2a^2+ab)\geq\sum_{cyc}a\sqrt{5b^2+4c^2},$$ where $a$, $b$ and $c$ are non-negatives.
Indeed, after squaring of the both sides we need to prove that: $$\sum_{cyc}(4a^4+8a^2b^2+a^2b^2+2a^2bc+4a^3b+4a^3c+4a^2bc-9a^2b^2)\geq2\sum_{cyc}ab\sqrt{5b^2+4c^2)(5c^2+4a^2)}$$ or $$\sum_{cyc}(2a^4+2a^3b+2a^3c+3a^2bc)\geq\sum_{cyc}ab\sqrt{5b^2+4c^2)(5c^2+4a^2)}$$ or $$(a+b+c)(2(a^3+b^3+c^3)+3abc)\geq\sum_{cyc}ab\sqrt{5b^2+4c^2)(5c^2+4a^2)}.$$ Now, by C-S and Rearrangement $$\sum_{cyc}ab\sqrt{5b^2+4c^2)(5c^2+4a^2)}\leq\sqrt{\sum_{cyc}ab\sum_{cyc}ab(5b^2+4c^2)(5c^2+4a^2)}=$$ $$=\sqrt{\sum_{cyc}ab\sum_{cyc}(20a^3b^3+20a^4bc+25a^3b^2c+16a^3c^2b)}\leq$$ $$\leq\sqrt{\sum_{cyc}ab\sum_{cyc}(20a^3b^3+29a^4bc+16a^3b^2c+16a^3c^2b)}.$$ Id est, it's enough to prove that: $$(a+b+c)^2(2(a^3+b^3+c^3)+3abc)^2\geq(ab+ac+bc)\sum_{cyc}(20a^3b^3+29a^4bc+16a^3b^2c+16a^3c^2b).$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$, $abc=w^3$ and $u^2=tv^2.$
Thus, $t\geq1$ and we need to prove that: $$9u^2(54u^3-54yv^2+6w^3+3w^3)^2\geq$$ $$\geq3v^2(20(27v^6-27uv^2w^3+3w^6)+29(27u^3w^3-27uv^2w^3+3w^6)+16(9uv^2-3w^3)w^3)$$ or $$27u^2(6u^3-6uv^2+w^3)^2\geq v^2(60v^6+87u^3w^3-131uv^2w^3+11w^6)$$ or $$972u^8-1944u^6v^2+972u^4v^4-60v^8+(324u^5-411u^3v^2+131uv^4)w^3+(27u^2-11v^2)w^6\geq0,$$ which is true by Schur ($w^3\geq4uv^2-3u^3$) twice: $$972u^8-1944u^6v^2+972u^4v^4-60v^8+(324u^5-411u^3v^2+131uv^4)w^3+(27u^2-11v^2)w^6\geq$$ $$\geq972u^8-1944u^6v^2+972u^4v^4-60v^8+(324u^5-411u^3v^2+131uv^4)w^3+(27u^2-11v^2)(4uv^2-3u^3)w^3=$$ $$=972u^8-1944u^6v^2+972u^4v^4-60v^8+(243u^5-270u^3v^2+87uv^4)w^3\geq$$ $$\geq972u^8-1944u^6v^2+972u^4v^4-60v^8+(243u^5-270u^3v^2+87uv^4)(4uv^2-3u^3)=$$ $$=3v^8t(t-1)(81t^3+27t^2-96t+20)\geq0.$$