# Is it true that $\frac{a_1^2}{a_2}+\frac{a_2^2}{a_3}+\cdots+\frac{a_n^2}{a_1}\geq\sqrt{n(a_1^2+a_2^2+\cdots+a_n^2)}$ for positive real $a_i$?

This problem is my attempt to generalize an easier problem:

Let $$a_1,a_2,\dots,a_n$$ be $$n$$ positive real numbers. Prove or disprove that: $$\frac{a_1^2}{a_2} + \frac{a_2^2}{a_3} + \dots + \frac{a_n^2}{a_1} \geq \sqrt{n \left( a_1^2 + a_2^2 +\cdots+a_n^2 \right)}.$$

I was able to prove that the inequality holds for $$n \leq 4$$ using AM-GM and Cauchy-Schwarz but was stuck with $$n \geq 5$$ after much effort. I don't know if this generalization is true and if it really is then it'll be great if someone can show me some hints to prove it.

Edit: To show my effort, I have included my solution to the case $$n=4$$ (for $$n \leq 3$$, a similar method can be applied)

For $$n=4$$, we need to prove that: $$\frac{a_1^2}{a_2} + \frac{a_2^2}{a_3} + \frac{a_3^2}{a_4} + \frac{a_4^2}{a_1} \geq \sqrt{4 \left(a_1^2 +a_2^2 + a_3^2 + a_4^2 \right)}$$

By using the Cauchy-Schwarz inequality, we have: $$\left( a_1^2a_2 + a_2^2a_3 + a_3^2a_4 + a_4^2a_1 \right) \left( \frac{a_1^2}{a_2} + \frac{a_2^2}{a_3} + \frac{a_3^2}{a_4} + \frac{a_4^2}{a_1} \right) \geq \left( a_1^2 + a_2^2 + a_3^2 + a_4^2 \right)^2$$

We also have: $$a_1^2a_2 + a_2^2a_3 + a_3^2a_4 + a_4^2a_1 \leq \sqrt{\left( a_1^2 + a_2^2 + a_3^2 + a_4^2 \right) \left( a_1^2a_2^2 + a_2^2a_3^2 + a_3^2a_4^2 + a_4^2a_1^2 \right) }$$

Note that: $$a_1^2a_2^2 + a_2^2a_3^2 + a_3^2a_4^2 + a_4^2a_1^2 = \left( a_1^2 + a_3^2 \right) \left( a_2^2 + a_4^2 \right) \leq \frac{\left( a_1^2+a_2^2+a_3^2+a_4^2 \right) ^2}{4}$$

$$\Longrightarrow \sqrt{\left( a_1^2 + a_2^2 + a_3^2 + a_4^2 \right) \left( a_1^2a_2^2 + a_2^2a_3^2 + a_3^2a_4^2 + a_4^2a_1^2 \right) } \leq \sqrt{\frac{\left( a_1^2 + a_2^2 + a_3^2 + a_4^2 \right)^3}{4}}$$

Therefore: $$\frac{a_1^2}{a_2} + \frac{a_2^2}{a_3} + \frac{a_3^2}{a_4} + \frac{a_4^2}{a_1} \geq \frac{\left( a_1^2 + a_2^2 + a_3^2 + a_4^2 \right)^2}{\sqrt{\dfrac{\left( a_1^2 + a_2^2 + a_3^2 + a_4^2 \right)^3}{4}}} = \sqrt{4 \left(a_1^2 +a_2^2 + a_3^2 + a_4^2 \right)} \text{ (QED)}$$

• Welcome to MSE. Remember to include your work on the problem. In this case, for example, could you share your proof for $n \le 4$. If you show your work your question will be better received in this site and will increase your chances of getting help. Jul 5, 2021 at 11:24

It's wrong for $$n=10$$.

The counterexample see here (the last post): https://artofproblemsolving.com/community/c6h314322p2091203

For $$n=6$$ you can see a stronger one here https://artofproblemsolving.com/community/c6h366576

Remarks: 1. This conjecture is No. 4 of Mr. Xuezhi Yang's 22 conjectures proposed at the Symposium in Elementary Mathematics (in 2009, China).

1. The inequality does not hold for $$n \ge 9$$, according to the literature.

2. I saw a proof for $$n = 5$$ in Mr. Xuezhi Yang's book (written in Chinese). It was written in the book that the inequality for $$n = 5$$ was proposed in July 07, 2003. I put it here (I reworte it accordingly. )

Since $$\frac{u^2}{v} = 2u - v + \frac{(u - v)^2}{v}$$, it suffices to prove that $$\sum_{\mathrm{cyc}} a_1 + \sum_{\mathrm{cyc}} \frac{(a_1 - a_2)^2}{a_2} \ge \sqrt{5 \sum_{\mathrm{cyc}} a_1^2}$$ or $$\left(\sum_{\mathrm{cyc}} a_1 + \sum_{\mathrm{cyc}} \frac{(a_1 - a_2)^2}{a_2}\right)^2 \ge 5 \sum_{\mathrm{cyc}} a_1^2$$ or $$\left(\sum_{\mathrm{cyc}} \frac{(a_1 - a_2)^2}{a_2}\right)^2 + 2 \sum_{\mathrm{cyc}} a_1 \cdot \sum_{\mathrm{cyc}} \frac{(a_1 - a_2)^2}{a_2} - \sum_{1\le i < j\le 5} (a_i - a_j)^2 \ge 0$$ where we have used $$5 \sum_{\mathrm{cyc}} a_1^2 - (\sum_{\mathrm{cyc}} a_1)^2 = \sum_{1\le i < j\le 5} (a_i - a_j)^2$$.

By Cauchy-Bunyakovsky-Schwarz inequality, we have $$\sum_{\mathrm{cyc}} a_1 \cdot \sum_{\mathrm{cyc}} \frac{(a_1 - a_2)^2}{a_2} \ge \left(\sum_{\mathrm{cyc}} |a_1 - a_2|\right)^2.$$ It suffices to prove that $$2\left(\sum_{\mathrm{cyc}} |a_1 - a_2|\right)^2 - \sum_{1\le i < j\le 5} (a_i - a_j)^2 \ge 0.$$ It suffices to prove that $$2\sum_{\mathrm{cyc}} (a_1 - a_2)^2 + 4 \sum_{\mathrm{cyc}} |(a_1 - a_2)(a_3 - a_4)| - \sum_{1\le i < j\le 5} (a_i - a_j)^2 \ge 0$$ or $$4 \sum_{\mathrm{cyc}} |(a_1 - a_2)(a_3 - a_4)| + 2 \sum_{\mathrm{cyc}} (a_1 - a_2)(a_3 - a_4) \ge 0 \tag{1}$$ which is clearly true. To obtain (1), we have used $$2\sum_{\mathrm{cyc}} (a_1 - a_2)^2 - \sum_{1\le i < j\le 5} (a_i - a_j)^2 = 2 \sum_{\mathrm{cyc}} (a_1 - a_2)(a_3 - a_4).$$

We are done.