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.
 A: Here is my proof
by AM-GM inequality,we have
$$ a\sqrt[3]{a+b}=\frac{3\sqrt[3]{2}a(a+b)}{3\sqrt[3]{2(a+b)(a+b)}}\geq 3\sqrt[3]{2}\cdot \frac{a(a+b)}{2+2a+2b} $$
Thus,it's suffice to prove that
$$ \frac{a(a+b)}{a+b+1}+\frac{b(b+c)}{b+c+1}+\frac{c(c+a)}{c+a+1}\geq 2 $$
Or
$$ \frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\leq 1 $$
After homogenous,it's 
$$\frac{a}{4a+4b+c}+\frac{b}{4b+4c+a}+\frac{c}{4c+4a+b}\leq \frac{1}{3} $$
Now,multiply $4a+4b+4c$ to each sides.we can rewrite the inequality into
$$ \frac{9ca}{4a+4b+c}+\frac{9ab}{4b+4c+a}+\frac{9bc}{4c+4a+b}\leq a+b+c $$
Using Cauchy-Schwarz inequality,we have
$$ \frac{9}{4a+4b+c}=\frac{(2+1)^2}{2(2a+b)+(2b+c)}\le \frac{2}{2a+b}+\frac{1}{2b+c} $$
Therefore
\begin{align}
  \sum{\frac{9ca}{4a+4b+c}}&\leq \sum{\left(\frac{2ca}{2a+b}+\frac{ca}{2b+c}\right)}\\
&=a+b+c 
\end{align}
Hence we are done!
A: 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 $\nabla f=0$.
