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 $a_i$

algebra-precalculusconjecturesexamples-counterexamplesinequality

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

Best Answer

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

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