Question from a Moscow summer camp Prove that, given $a,b,c > 0$ and $n$ a positive integer, $$\frac{a^n}{b+c}+\frac{b^n}{a+c}+\frac{c^n}{a+b} \geq \frac{a^{n-1}+b^{n-1}+c^{n-1}}{2}\ .$$ I've tried every rathole for hours on this little fellow... It seems simple, and can be intuited, but it‘s devilish to solve. A good first step (I don't want to give out my four or five approaches for fear of distracting), I think, is to let $a \geq b \geq c$. Thanks for any help!

$$\sum_{cyc}\frac{a^n}{b+c}-\frac{a^{n-1}+b^{n-1}+c^{n-1}}{2}=\sum_{cyc}\left(\frac{a^n}{b+c}-\frac{a^{n-1}}{2}\right)=$$ $$=\sum_{cyc}\frac{a^{n-1}(a-b-(c-a))}{2(b+c)}=\sum_{cyc}(a-b)\left(\frac{a^{n-1}}{2(b+c)}-\frac{b^{n-1}}{2(c+a)}\right)=$$ $$=\sum_{cyc}\frac{(a-b)(c(a^{n-1}-b^{n-1})+a^n-b^n)}{2(a+c)(b+c)}\geq0$$ because $$(a-b)(a^{n-1}-b^{n-1})\geq0$$ and $$(a-b)(a^n-b^n)\geq0.$$