Prove that $\sum_{cyc} \sqrt[n]{\frac{a}{b+c}} \ge \frac{3}{\sqrt[n]{2}}$ Given a,b,c are positive real numbers and n is positive integer number.
Prove that
$$\sum_{cyc} \sqrt[n]{\frac{a}{b+c}} \ge \frac{3}{\sqrt[n]{2}}$$
I tried expanding it, using equivalence transformation, using inequalities AM-GM, Holder.., but no success. I can only prove the above inequality for n=1
$$\sum_{cyc} \frac{a}{b+c}=\sum_{cyc} \frac{a^2}{ab+ac}\geq \frac{(a+b+c)^2}{2(ab+bc+ca)}\geq \frac{3(ab+bc+ca)}{2(ab+bc+ca)}=\frac{3}{2}$$
I hope your help, thanks

Your conjecture does not hold for $n > 1$. Let $a = b > 0$ and $c \to  0$ then, for $n > 1$,
$$
\sum_{cyc} \sqrt[n]{\frac{a}{b+c}} \to  2 \le \frac{3}{\sqrt[n]{2}}
$$
Indeed, $2$ is a tight lower limit for $n > 1$.
