<s>[INST] How prove this $a+b+c+3\sqrt[3]{abc}\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})$ 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?
 [/INST] 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 $(*)$.
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