Problem 2143

Let $S=\sum_{k=1}^{100} a_{k}^{2} a_{k+1}$. (As usual, we consider the indices modulo $100$, e.g. we set $a_{101}=a_{1}$ and $a_{102}=a_{2}$.)
Applying the Cauchy-Schwarz inequality to sequences $\left(a_{k+1}\right)$ and $\left(a_{k}^{2}+2 a_{k+1} a_{k+2}\right)$, and then the AM-GM inequality to numbers $a_{k+1}^{2}$ and $a_{k+2}^{2}$,
$$\begin{align*} (3 S)^{2} & =\left(\sum_{k=1}^{100} a_{k+1}\left(a_{k}^{2}+2 a_{k+1} a_{k+2}\right)\right)^{2} \leq\left(\sum_{k=1}^{100} a_{k+1}^{2}\right)\left(\sum_{k=1}^{100}\left(a_{k}^{2}+2 a_{k+1} a_{k+2}\right)^{2}\right) \\ & =1 \cdot \sum_{k=1}^{100}\left(a_{k}^{2}+2 a_{k+1} a_{k+2}\right)^{2}=\sum_{k=1}^{100}\left(a_{k}^{4}+4 a_{k}^{2} a_{k+1} a_{k+2}+4 a_{k+1}^{2} a_{k+2}^{2}\right) \\ & \leq \sum_{k=1}^{100}\left(a_{k}^{4}+2 a_{k}^{2}\left(a_{k+1}^{2}+a_{k+2}^{2}\right)+4 a_{k+1}^{2} a_{k+2}^{2}\right)=\sum_{k=1}^{100}\left(a_{k}^{4}+6 a_{k}^{2} a_{k+1}^{2}+2 a_{k}^{2} a_{k+2}^{2}\right) . \end{align*}$$
Applying the trivial estimates
$$\sum_{k=1}^{100}\left(a_{k}^{4}+2 a_{k}^{2} a_{k+1}^{2}+2 a_{k}^{2} a_{k+2}^{2}\right) \leq\left(\sum_{k=1}^{100} a_{k}^{2}\right)^{2} \quad \text{ and } \quad \sum_{k=1}^{100} a_{k}^{2} a_{k+1}^{2} \leq\left(\sum_{i=1}^{50} a_{2 i-1}^{2}\right)\left(\sum_{j=1}^{50} a_{2 j}^{2}\right)$$
we obtain that
$$(3 S)^{2} \leq\left(\sum_{k=1}^{100} a_{k}^{2}\right)^{2}+4\left(\sum_{i=1}^{50} a_{2 i-1}^{2}\right)\left(\sum_{j=1}^{50} a_{2 j}^{2}\right) \leq 1+\left(\sum_{i=1}^{50} a_{2 i-1}^{2}+\sum_{j=1}^{50} a_{2 j}^{2}\right)^{2}=2,$$
hence
$$S \leq \frac{\sqrt{2}}{3} \approx 0.4714<\frac{12}{25}=0.48$$