For positive numbers $a$, $b$, $c \geq 0$ and $a+b+c=1$ show that:
$\sqrt[3]{4+17a^2b}+\sqrt[3]{4+17b^2c}+\sqrt[3]{4+17c^2a}+10 \Big(\frac{1}{27}-abc \Big) \geq 5$
I tried to use $AM-GM$ with $\frac{17}{27}+17a^2b$ and tried after that use Holder, but after that stucked.
Cubing both sides doesn't work too.
Best Answer
Just AM-GM and Holder help here!
Let $abc=\frac{x}{27}$.
Thus, by AM-GM $$abc\leq\left(\frac{a+b+c}{3}\right)^3=\frac{1}{27}$$ and by AM-GM and Holder we obtain: $$\sum_{cyc}\sqrt[3]{4+17a^2b}\geq3\sqrt[9]{\prod_{cyc}(4+17a^2b)}\geq3\sqrt[3]{4+17abc}=3\sqrt[3]{4+\frac{17x}{27}}=\sqrt[3]{108+17x}$$ and it's enough to prove that $$\sqrt[3]{108+17x}+10\left(\frac{1}{27}-\frac{x}{27}\right)\geq5$$ or $$27\sqrt[3]{108+17x}\geq125+10x,$$ where $0\leq x\leq1,$ which is true because $$27\sqrt[3]{108+17x}=9\sqrt[3]{27\cdot108+27\cdot17x}\geq9\sqrt[3]{(14+x)^3}=9(14+x)\geq125+10x.$$