# Find the smallest value of $\sqrt[5]{\frac{abc}{b+c}} + \sqrt[5]{\frac{b}{c(1+ab)}} + \sqrt[5]{\frac{c}{b(1+ac)}}$

Let $$a\ge0$$ and $$b,c>0$$, we need to find the smallest value of the expression

$$S=\sqrt[5]{\frac{abc}{b+c}} + \sqrt[5]{\frac{b}{c(1+ab)}} + \sqrt[5]{\frac{c}{b(1+ac)}}$$

I have no idea for this question, does anyone can help me to answer this?

• Hmm, setting $a=0$ gives you something symmetric in $b$ and $c$ therefore $2$ seems to be the smallest value that way. – Ali Caglayan Sep 7 '15 at 6:06

Hint:

1. Let $x = a, y = \frac1b, z = \frac1c$

2. Suppose $x + y + z = 1$

3. Study the function $x \mapsto \left(\frac{x}{1 - x}\right)^{\frac15}$ for $0 < x < 1$.

• Very nice symmetrisation via steps 1 & 2. However, how does step 3 help? – Macavity Sep 8 '15 at 4:25

For $a=0$ and $b=c=1$ we have a value $2$.

We'll prove that it's a minimal value.

Indeed, let $a\neq0$, $a=x$, $b=\frac{1}{y}$ and $c=\frac{1}{z}$.

Hence, by AM-GM we obtain: $$S=\sum_{cyc}\sqrt[5]{\frac{x}{y+z}}\geq\sum_{cyc}\sqrt{\frac{x}{y+z}}=$$ $$=\sum_{cyc}\frac{2x}{2\sqrt{x(y+z)}}\geq\sum_{cyc}\frac{2x}{x+y+z}=2$$ and we are done!