How prove this $a+b+c+3\sqrt[3]{abc}\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})$

Question

let $a,b,c>0$, show that $$a+b+c+3\sqrt[3]{abc}\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})$$

I know this $$a+b+c\ge 3\sqrt[3]{abc}$$ so $$\Longleftrightarrow 6\sqrt[3]{abc}\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})$$ But this maybe not true?

Answer 1

WLOG suppose that $a\geq b\geq c$.

First see that using weighted AG-inequality, we have: $$ c+3\sqrt[3]{abc}\geq 4\sqrt c \sqrt[4]{ab} (*) $$ Then we have: $$ a+b+c+3\sqrt[3]{abc}- 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})=\\ (\sqrt{a}+\sqrt{b}-\sqrt{c})^2-4\sqrt{ab}+3\sqrt[3]{abc}\\ \geq (2\sqrt[4]{ab}-\sqrt{c})^2-4\sqrt{ab}+3\sqrt[3]{abc}\\ = c- 4\sqrt c \sqrt[4]{ab}+3\sqrt[3]{abc}\geq 0\\ $$ The last one is positive due to $(*)$.

Answer 2

Let $a=x^6$, $b=y^6$ and $c=z^6$, where $x$, $y$ and $z$ be positive numbers.

Hence, by Schur and AM-GM we obtain: $$a+b+c+3\sqrt[3]{abc}=\sum_{cyc}(x^6+x^2y^2z^2)\geq\sum_{cyc}(x^4y^2+x^4z^2)=$$ $$=\sum_{cyc}(x^4y^2+y^4x^2)\geq2\sum_{cyc}x^3y^3=2(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}).$$ Done!