If $abc=1$, then $\frac{a^{n+2}}{a^n+(n-1)b^n}+\frac{b^{n+2}}{b^n+(n-1)c^n}+\frac{c^{n+2}}{c^n+(n-1)a^n} \geq \frac{3}{n} $ Let $a$, $b$ and $c$ be positive real numbers with $abc=1$. Prove that
  $$
\frac{a^{n+2}}{a^n+(n-1)b^n}+\frac{b^{n+2}}{b^n+(n-1)c^n}+\frac{c^{n+2}}{c^n+(n-1)a^n} \geq \frac{3}{n}
$$
  for each integer $n$.


I have used Cauchy-Schwarz inequalities and Jensen inequality. But I am stuck. I need some idea and advice on this problem. Induction would be cruel.
Using AM-GM we get (hereon $\sum$ denotes cyclic sums): $$\sum \frac{a^{n+2}}{a^n+(n-1)b^n} =\sum \left( a^2- (n-1)\frac{a^2b^n}{a^n+(n-1)b^n}\right) \\ \ge \sum\left( a^2- (n-1)\frac{a^2b^n}{n \cdot a\cdot b^{n-1}}\right)= \sum a^2-\frac{n-1}n\sum ab$$

So it is enough to show that 
$$n \sum a^2 \ge (n-1)\sum ab+3$$
which follows from $\sum a^2 \ge \sum ab$ and $\sum a^2 \ge 3$ by AM-GM.
You can see that every term has value $\frac{1}{n}$ in case $a = b = c = 1$.
Try to prove that changing one of the variables (let $a &gt; 1$,  $b &lt; 1$,  $c &lt; 1$ be the first case and $a &gt; 1$, $b &gt; 1$, $c &lt; 1$ - the second one and $a &gt; 1$, $b &lt; 1$, $c &gt; 1$ - the third one) increases the total sum.