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