Prove that: $\sqrt[3]{a+2b}+\sqrt[3]{b+2c}+\sqrt[3]{c+2a}\le 3\sqrt[3]{3}$ Given $a,b,c&gt;0$ 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 $x = a+2b$, $y = b+2c$, and $z = c + 2a$, then $x+y+z = 3(a+b+c) = 9$, and we have that $f''(x) = -\dfrac{2}{9}\cdot x^{-\frac{5}{3}} &lt; 0$ on $(0,9)$ with $f(x) = \sqrt[3]{x}$. Thus $f$ is concave and we have:

$LHS = f(x) + f(y) + f(z) \leq 3f\left(\frac{x+y+z}{3}\right) = 3f(3) = 3\sqrt[3]{3} = RHS$
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
$$f(a,b,c)=\sqrt[3]{a+2b}+\sqrt[3]{b+2c}+\sqrt[3]{c+2a}\le3\sqrt[3]{3}$$
$$\sqrt[3]{a+2b}+\sqrt[3]{b+2c}+\sqrt[3]{c+2a}\le3\sqrt[3]{3}$$