Given $n$ positive numbers $x_1,\ldots,x_n$ ($n\ge 3$) such that the product $x_1x_2\cdots x_n=1$, show that

$$\dfrac{x_1^8}{(x_1^4+x_2^4)x_2}+\frac{x_2^8}{(x_2^4+x_3^4)x_3}+\cdots+\frac{x_n^8}{(x_n^4+x_1^4)x_1}\ge \frac{n}{2}$$

I am having trouble demonstrating this for even $n=3$, but it seems that once that case is established the method for the general case should be identical. I am sure there is some nice application of AM-GM or Cauchy that will solve it. Notice that in each term we have the numerator of degree $8$, and in the denominator there is one factor of degree $4$ and also a linear factor which is quite an annoyance. I have tried to "homogenize" the linear factor so that it is of degree 4 (thus each fraction of degree 0) by doing some substitutions, but I have been unsuccessful with this.

If anyone can provide a hint (e.g suggest a substitution that may be helpful), I would greatly appreciate it. This problem has been bothering me for a while now.

You can prove it using the inequalities AM-GM, GM-HM (geometric mean and harmonic mean) and the inequality of ordered sequences: $$\sum_{cyc} \frac{x_1^8}{\left(x_1^4+x_2^4\right)x_2}=\sum_{cyc} \frac{x_1^4}{x_2}\left(\frac{x_1^4}{x_1^4+x_2^4}\right)=\sum_{cyc} \frac{x_1^4}{x_2}\left(1-\frac{x_2^4}{x_1^4+x_2^4}\right)=\sum_{cyc} \frac{x_1^4}{x_2}-\frac{1}{x_2}\frac{x_1^4x_2^4}{\left(x_1^4+x_2^4\right)}=\sum_{cyc} \frac{x_1^4}{x_2}-\frac{1}{2x_2}\frac{2}{\left(\frac{1}{x_1^4}+\frac{1}{x_2^4}\right)}=\sum_{cyc} \frac{x_1^4}{x_2}-\sum_{cyc} \frac{1}{2x_2}\frac{2}{\left(\frac{1}{x_1^4}+\frac{1}{x_2^4}\right)}$$
Due to GM-HM we have: $\frac{2}{\left(\frac{1}{x_1^4}+\frac{1}{x_2^4}\right)}\le x_1^2x_2^2$ and since $x_1^4, x_2^4,\dots, x_n^4$ and $\frac{1}{x_1},\dots,\frac{1}{x_n}$ are oppositely ordered, we have: $\sum_{cyc} \frac{x_1^4}{x_2}\ge\sum_{cyc} \frac{x_1^4}{x_1}=\sum_{cyc} x_1^3$. Therefore: $$\sum_{cyc} \frac{x_1^4}{x_2}-\sum_{cyc} \frac{1}{2x_2}\frac{2}{\left(\frac{1}{x_1^4}+\frac{1}{x_2^4}\right)}\ge\sum_{cyc} x_1^3-\sum_{cyc} \frac{1}{2x_2}x_1^2x_2^2=\sum_{cyc} x_1^3-\sum_{cyc} \frac{x_1^2x_2}{2}$$ So it is left to prove that: $$\sum_{cyc} x_1^3-\sum_{cyc} \frac{x_1^2x_2}{2}\ge \frac{n}{2}\iff 2\sum_{cyc} x_1^3\ge n+\sum_{cyc} x_1^2x_2$$ But due to AM-GM we have $\sum_{cyc} x_1^3\ge n\cdot\left(x_1^3x_2^3\cdots x_n^3\right)^{\frac{1}{n}}=n$ and $\sum_{cyc} x_1^3=\sum_{cyc} \frac{2x_1^3+x_2^3}{3}\ge\sum_{cyc} x_1^2x_2$ and adding these two yields the desired result.
• Very nice! +1. You can cut down a lot of steps by using just AM-GM, then rearrangements and finally an AM-GM: $$\sum_{cyc} \frac{x_1^4}{x_2}-\frac1{x_2}\frac{x_1^4x_2^4}{(x_1^4+x_2^4)} \ge \sum_{cyc} \frac{x_1^4}{x_2}-\frac1{x_2}\frac{x_1^4x_2^4}{2x_1^2x_2^2}= \sum_{cyc} \frac{x_1^4}{x_2}-\frac{x_1^2x_2}2 \ge \sum_{cyc} x_1^3-\frac{x_1^3}2 \ge \frac{n}2$$ – Macavity Mar 1 '15 at 12:15