Let $a_1,...,a_{100}$ be non-negative numbers such that $a_1^2+...+a_{100}^2=1$ Prove that $a_1^2.a_2+...+a_{100}^2.a_1\le \frac{12}{25}$

Question

Let $a_1,...,a_{100}$ be non-negative numbers such that $a_1^2+...+a_{100}^2=1$ Prove that $a_1^2.a_2+...+a_{100}^2.a_1\le \frac{12}{25}$

I was thinking about Cauchy Schwarz, but $4$ th powers make it worse.

Answer 1

Let $a_{101}=a_1$, $a_0=a_{100}$ and $a_{102}=a_2$. Hence, by C-S(cauchy -schwarz)(1) and AM-GM(2)

$\left(3\sum\limits_{i=1}^{100}a_i^2a_{i+1}\right)^2=\left(\sum\limits_{i=1}^{100}a_i\left(a_{i-1}^2+2a_ia_{i+1}\right)\right)^2\leq\sum\limits_{i=1}^{100}a_i^2\sum\limits_{i=1}^{100}\left(a_{i-1}^2+2a_ia_{i+1}\right)^2\tag1=$

$=\sum\limits_{i=1}^{100}\left(a_i^4+4a_i^2a_{i+1}a_{i+2}+4a_{i+1}^2a_{i+2}^2\right)\leq\sum\limits_{i=1}^{100}\left(a_i^4+2a_i^2\left(a_{i+1}^2+a_{i+2}^2\right)+4a_{i+1}^2a_{i+2}^2\right)=$

$=\sum\limits_{i=1}^{100}\left(a_i^4+2a_i^2a_{i+1}^2+2a_i^2a_{i+2}^2\right)+4\sum\limits_{i=1}^{100}a_{i+1}^2a_{i+2}^2\leq\left(\sum\limits_{i=1}^{100}a_i^2\right)^2+\left(\sum\limits_{i=1}^{100}a_i^2\right)^2=2\tag2$

Id est, $\sum\limits_{i=1}^{100}a_i^2a_{i+1}\leq\frac{\sqrt2}{3}<\frac{12}{25}$