Prove that: $\sqrt[3]{a+2b}+\sqrt[3]{b+2c}+\sqrt[3]{c+2a}\le 3\sqrt[3]{3}$ Given $a,b,c>0$ and $a+b+c=3$. Prove that: $\sqrt[3]{a+2b}+\sqrt[3]{b+2c}+\sqrt[3]{c+2a}\le 3\sqrt[3]{3}$

let $$f(a,b,c)=\sqrt[3]{a+2b}+\sqrt[3]{b+2c}+\sqrt[3]{c+2a}\tag{1}$$
and $$g(a,b,c)=a+b+c-3=0\tag{2}$$
Using Lagarange's Multiplier we have
$$\frac{\frac{\partial f}{\partial a}}{\frac{\partial g}{\partial a}}=
\frac{\frac{\partial f}{\partial b}}{\frac{\partial g}{\partial b}}=
\frac{\frac{\partial f}{\partial c}}{\frac{\partial g}{\partial c}}=k$$
This implies
$$(a+2b)^{-2/3}+2\cdot(c+2a)^{-2/3}=3k \tag{3}$$
$$(b+2c)^{-2/3}+2\cdot(a+2b)^{-2/3}=3k \tag{4}$$
$$(c+2a)^{-2/3}+2\cdot(2+2c)^{-2/3}=3k \tag{5}$$
Solving $(3),(4),(5)$ Simultaneously We get
$$a=b=c$$substituting this in $(2)$ we get,
$$a=b=c=1$$
It can be shown that Maxima occurs at above mentioned values of parameters,
Putting these values in $(1)$ we get
$$f(1,1,1)=\sqrt[3]{3}+\sqrt[3]{3}+\sqrt[3]{3}=3\sqrt[3]{3}$$
Since $f(1,1,1)=3\sqrt[3]{3}$ is Maximum values of $f(a,b,c)$ this implies
$$f(a,b,c)=\sqrt[3]{a+2b}+\sqrt[3]{b+2c}+\sqrt[3]{c+2a}\le3\sqrt[3]{3}$$
$$\sqrt[3]{a+2b}+\sqrt[3]{b+2c}+\sqrt[3]{c+2a}\le3\sqrt[3]{3}$$
