48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions

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 $\left(a_k^2+2a_{k+1}{a}_{k+2}\right)$, and then the AM-GM inequality to numbers $a_{k+1}^2$ and $a_{k+2}^2$, $$\begin{array}{rl}(3S{)}^2& ={\left(\sum _{k=1}^{100}{a}_{k+1}\left(a_k^2+2a_{k+1}{a}_{k+2}\right)\right)}^2\le \left(\sum _{k=1}^{100}{a}_{k+1}^2\right)\left(\sum _{k=1}^{100}{\left(a_k^2+2a_{k+1}{a}_{k+2}\right)}^2\right)\\ & =1\cdot \sum _{k=1}^{100}{\left(a_k^2+2a_{k+1}{a}_{k+2}\right)}^2=\sum _{k=1}^{100}\left(a_k^4+4a_k^2{a}_{k+1}{a}_{k+2}+4a_{k+1}^2{a}_{k+2}^2\right)\\ & \le \sum _{k=1}^{100}\left(a_k^4+2a_k^2\left(a_{k+1}^2+a_{k+2}^2\right)+4a_{k+1}^2{a}_{k+2}^2\right)=\sum _{k=1}^{100}\left(a_k^4+6a_k^2{a}_{k+1}^2+2a_k^2{a}_{k+2}^2\right).\end{array}$$ Applying the trivial estimates $$\sum _{k=1}^{100}\left(a_k^4+2a_k^2{a}_{k+1}^2+2a_k^2{a}_{k+2}^2\right)\le {\left(\sum _{k=1}^{100}{a}_k^2\right)}^2\text{ and }\sum _{k=1}^{100}{a}_k^2{a}_{k+1}^2\le \left(\sum _{i=1}^{50}{a}_{2i-1}^2\right)\left(\sum _{j=1}^{50}{a}_{2j}^2\right)$$ we obtain that $$(3S{)}^2\le {\left(\sum _{k=1}^{100}{a}_k^2\right)}^2+4\left(\sum _{i=1}^{50}{a}_{2i-1}^2\right)\left(\sum _{j=1}^{50}{a}_{2j}^2\right)\le 1+{\left(\sum _{i=1}^{50}{a}_{2i-1}^2+\sum _{j=1}^{50}{a}_{2j}^2\right)}^2=2,$$ hence $$S\le \frac{\sqrt{2}}{3}\approx 0.4714<\frac{12}{25}=0.48$$