Prove $a\sqrt[3]{a+b}+b\sqrt[3]{b+c}+c\sqrt[3]{c+a} \ge 3 \sqrt[3]2$ Prove $a\sqrt[3]{a+b}+b\sqrt[3]{b+c}+c\sqrt[3]{c+a} \ge 3 \sqrt[3]2$ with $a + b+c=3 \land a,b,c\in \mathbb{R^+}$

I tried power mean inequalities but I still can't prove it.

Hint, you can try to use calculus. Let $$f(a, b, c, \lambda )=a\sqrt[3]{a+b}+b\sqrt[3]{b+c}+c\sqrt[3]{c+a}-\lambda (a+b+c-3)=0.$$
Then, the extrema occurs at $\nabla f=0$.
