Let $a,b,c$ be nonnegative real numbers such that $a+b+c=3$. Prove that
\[ \sqrt[3]{3a^2+2ab+3b^2}+\sqrt[3]{3b^2+2bc+3c^2}+\sqrt[3]{3c^2+2ca+3a^2} \geq 6.\]

Proof：(by gxggs)

By Holder inequality, we have
\[\left(\sum_{cyc}{\sqrt[3]{3a^2+2ab+3b^2}}\right)^3\sum_{cyc}{\left(\frac{(a+b)^4}{3a^2+2ab+3b^2}\right)}\geq 16(a+b+c)^4\]
Therefore it suffices to show that
\[ 16(a+b+c)^4\geq 216\cdot \sum_{cyc}{\left(\frac{(a+b)^4}{3a^2+2ab+3b^2}\right)}\]
Or
\[2(a+b+c)^2\geq 3\cdot \sum_{cyc}{\left(\frac{(a+b)^4}{3a^2+2ab+3b^2}\right)}\]
\[\Longleftrightarrow \sum_{cyc}{\left(a^2+b^2+4ab-\frac{3(a+b)^4}{3a^2+2ab+3b^2}\right)}\geq 0\]
which is equivalent to
\[\frac{2ab(a-b)^2}{3a^2+2ab+3b^2}+\frac{2bc(b-c)^2}{3b^2+2bc+3c^2}+\frac{2ca(c-a)^2}{3c^2+2ca+3a^2}\geq 0\]
which is obvious true. Equal holds when $a=b=c=1$ or $a=3,b=0,c=0$ or their cyclic forms.   $\blacksquare$
另外，越南TaHongQuang 作了如下推广

 1.Let $a, \, b, \, c > 0$ such that $a+b+c =\frac{1}{4}$ .  Prove that
\[  \sqrt [3]{65{a}^{2}+81{b}^{2}-2ab}+\sqrt [3]{65{b}^{2}+81{c}^{2}-2bc}+\sqrt [3]{65{c}^{2}+81{a}^{2}-2ac} \ge 3 \]

2.Let $a, \, b, \, c > 0$ such that $a+b+c =\frac{1}{6}$ .  Prove that
\[  \sqrt [3]{7{a}^{2}+9{b}^{2}-4ba}+\sqrt [3]{7{b}^{2}+9{c}^{2}-4bc}+\sqrt [3]{7{c}^{2}+9{a}^{2}-4ac}\ge 1 \]