China National Team Selection Test

Firstly, we prove that $$\underset{1\le k\le n}{\max }{a}_k<2,\text{ and }\underset{1\le k\le n}{\max }{b}_k<\frac{2}{n-1}.$$ Let $L={\max }_{1\le k\le n}{a}_k$. From (b) and ${\sum }_{k=1}^n{b}_k=1$, we get $L^2\le 1+L$, so $L<2$. Let $b_m={\max }_{1\le k\le n}{b}_k$. Then by using $2b_k\ge b_{k-1}+b_{k+1}$, it is easy to see that $$b_k\ge \{\begin{array}{ll}\frac{(k-1)b_m+(m-k)b_1}{m-1},& 1\le k\le m,\\ \frac{(k-m)b_n+(n-k)b_m}{n-m},& m\le k\le n.\end{array}$$ Since $b_1>0$ and $b_m>0$, so $$b_k>\{\begin{array}{ll}\frac{k-1}{m-1}{b}_m,& 1\le k\le m,\\ \frac{n-k}{n-m}{b}_m,& m\le k\le n.\end{array}$$ It follows that $$\begin{array}{rl}1& =\sum _{k=1}^n{b}_k=\sum _{k=1}^m{b}_k+\sum _{k=m+1}^n{b}_k\\ & >\frac{1}{m-1}\left(\sum _{k=1}^m(k-1)\right)b_m+\frac{1}{n-m}\left(\sum _{k=m+1}^n(n-k)\right)b_m\end{array}$$ $$=\frac{m}{2}{b}_m+\frac{n-m-1}{2}{b}_m=\frac{n-1}{2}{b}_m.$$ So $b_m<\frac{2}{n-1}$, that is ${\max }_{1\le k\le n}{b}_k<\frac{2}{n-1}$.

Now let $f_0=1$, $f_k=1+{\sum }_{i=1}^k{a}_i{b}_i$, $k=1,2,\dots ,n$. Then $f_k-f_{k-1}=a_k{b}_k$, and from (b) we have $a_k^2\le f_k$, i.e. $a_k\le \sqrt{f_k}$, $k=1,2,\dots ,n$. Since ${\max }_{1\le k\le n}{a}_k<2$, so $$f_k-f_{k-1}=a_k{b}_k\le b_k\sqrt{f_k}$$ and $$f_k-f_{k-1}<2b_k.$$ Thus, for $1\le k\le n$, $$\begin{array}{rl}\sqrt{f_k}-\sqrt{f_{k-1}}& <b_k\cdot \frac{\sqrt{f_k}}{\sqrt{f_k}+\sqrt{f_{k-1}}}\\ & =b_k\left(\frac{1}{2}+\frac{f_k-f_{k-1}}{2(\sqrt{f_k}+\sqrt{f_{k-1}}{)}^2}\right)\\ & <b_k\left(\frac{1}{2}+\frac{2b_k}{2(\sqrt{f_k}+\sqrt{f_{k-1}}{)}^2}\right)\\ & <b_k\left(\frac{1}{2}+\frac{b_k}{4}\right)\\ & <\left(\frac{1}{2}+\frac{1}{2(n-1)}\right)b_k.\end{array}$$ Hence, summing from $k=1$ to $n$, $$\begin{array}{r}{a}_n\le \sqrt{f_n}<\sqrt{f_0}+\sum _{k=1}^n\left(\frac{1}{2}+\frac{1}{2(n-1)}\right)b_k\\ =\frac{3}{2}+\frac{1}{2(n-1)}.\end{array}$$ Let $n\to +\infty$, we obtain $a_n\le \frac{3}{2}$.

When $a_k=1+\frac{k}{2n}$, $b_k=\frac{1}{n}$, $k=1,2,\dots ,n$, we have $$a_k^2={\left(1+\frac{k}{2n}\right)}^2\le 1+\sum _{i=1}^k\frac{1}{n}\left(1+\frac{i}{2n}\right)$$ Hence the maximum value is $\frac{3}{2}$.