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} $

Question

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.

Answer 1

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.

Answer 2

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 $a > 1$, $b < 1$, $c < 1$ be the first case and $a > 1$, $b > 1$, $c < 1$ - the second one and $a > 1$, $b < 1$, $c > 1$ - the third one) increases the total sum.