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 $a_i$?

Question

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

Answer 1

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

Answer 2

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

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

3. 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. )

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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.