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?
2026-03-25 18:48:06.1774464486
How prove this $a+b+c+3\sqrt[3]{abc}\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})$
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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 $(*)$.