Problem. Let $a,b,c\ge 0: a+b+c=3.$ Prove that $$a\sqrt{b^2+c^2+abc}+b\sqrt{a^2+c^2+abc}+c\sqrt{b^2+a^2+abc}\ge 3\sqrt{3abc}.$$
I've tried to use AM-GM.
Firstly, we can rewrite the OP as$$\sum_{cyc}a\sqrt{1+\frac{b^2+c^2}{abc}}\ge 3\sqrt{3}.$$ We have $b^2+c^2\ge 2bc$ and it leads to wrong inequality$$a\sqrt{1+\frac{2}{a}}+b\sqrt{1+\frac{2}{b}}+c\sqrt{1+\frac{2}{c}}\ge 3\sqrt{3}.$$ Also, by AM-GM $$\sum_{cyc}a\sqrt{1+\frac{b^2+c^2}{abc}}\ge 3\sqrt[3]{abc\cdot\sqrt{\prod\limits_{cyc}\dfrac{b^2+c^2+abc}{abc}}}.$$ Hence, we need to prove $$(a^2+b^2+abc)(b^2+c^2+abc)(c^2+a^2+abc)\ge 27abc.$$ I used $$a^2+b^2+abc\ge 3\sqrt[3]{a^3b^3c},$$which gave reverse $abc\ge 1.$
Hope to see better ideas. Thank you.