# 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.

• Have you tried to invert both sides and calculate the left side? – Lehs Oct 8 '14 at 16:46
• Presumably for each positive integer $n$. It's certainly not true for $n=0$ – Robert Israel Oct 8 '14 at 17:10

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.

• @RonAld I have used $a^n + (n-1)b^n \ge n a b^{n-1}$, which follows from AM-GM, no need to assume any ordering. Taking reciprocals and multiplying by a negative numerator preserves the $\ge$ sign. – Macavity Oct 13 '14 at 11:13
• Is that mean that you use this? a^n+(n-1)b^n >= 2(a^n(n-1)b^n)^(1/2). I am really sorry but I am still confused why that is true. – Ron Ald Oct 13 '14 at 11:40
• yes yes. your right. i was so dumb. i am not familiar with cyclic. when you say Summation a^2. is that mean, a^2+b^2+c^2. and when you say Summation ab = ab+bc+ca. is that true? – Ron Ald Oct 13 '14 at 12:05
• Yes, that is correct both times. Sorry for using a notation you were not familiar with, but it may be useful to learn as you will find it used often. – Macavity Oct 13 '14 at 12:07
• god dammit. your a genius and I really meant it. How could it be so easy when i know the answer. I am dying from this problem. you saved my life. I tried so many many ways, still no help. WOW you are my savior. How do i add you as my friend here? – Ron Ald Oct 13 '14 at 12:11

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.

• we can assume that c<=b<=a, then case 1 c<=1<b<=a and case 2 c<=b<=1<=a. but there can't be case 3 right. because if you assume a>1, b<1 and c>1 then c>1>b which is not true because we have assumed in the first place that c<=b<=a. I have tried both cases but not really helping me. I need some more idea. – Ron Ald Oct 8 '14 at 17:45