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: $\sqrt[3]{a+2b}+\sqrt[3]{b+2c}+\sqrt[3]{c+2a}\le 3\sqrt[3]{3}$
 A: Let $x = a+2b$, $y = b+2c$, and $z = c + 2a$, then $x+y+z = 3(a+b+c) = 9$, and we have that $f''(x) = -\dfrac{2}{9}\cdot x^{-\frac{5}{3}} < 0$ on $(0,9)$ with $f(x) = \sqrt[3]{x}$. Thus $f$ is concave and we have:
$LHS = f(x) + f(y) + f(z) \leq 3f\left(\frac{x+y+z}{3}\right) = 3f(3) = 3\sqrt[3]{3} = RHS$
A: Another way is to use Power Means inequality
$$\left(\frac{\sqrt[3]{a+2b}+\sqrt[3]{b+2c}+\sqrt[3]{c+2a} }3\right)^3 \le \frac{(a+2b)+(b+2c)+(c+2a)}3$$
$$\implies LHS \le 3\sqrt[3]3$$
A: 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
$$f(a,b,c)=\sqrt[3]{a+2b}+\sqrt[3]{b+2c}+\sqrt[3]{c+2a}\le3\sqrt[3]{3}$$
$$\sqrt[3]{a+2b}+\sqrt[3]{b+2c}+\sqrt[3]{c+2a}\le3\sqrt[3]{3}$$
