Team Selection Test

The answer is $50$. If $a_0<0$ then it is clear that $a_{100}>a_0$. If $a_0=0$, then $b_n\in \{0,1/2\}$ and hence $S=b_1+b_2+\cdots +b_{100}\le 50$. Assume that $a_0>0$. In this case all terms of the sequence $(a_n)$ are positive. We have $$(a_n-\frac{a_{n-1}}{2})(a_n-2a_{n-1}^2)=0\text{and}{a}_{100}\le a_0.$$ This equation can be expressed as $$\frac{a_n}{a_{n-1}}+\frac{a_{n-1}^2}{a_n}=2a_{n-1}+\frac{1}{2}.$$

Summing up the equations for $n=1,2,...,100$ side by side, we get $$\sum _{n=1}^{100}\frac{a_n}{a_{n-1}}+\sum _{n=1}^{100}\frac{a_{n-1}^2}{a_n}=2\sum _{n=1}^{100}{a}_{n-1}+50.$$ Using Cauchy-Schwarz Inequality $$\sum _{n=1}^{100}\frac{a_{n-1}^2}{a_n}\ge \frac{(a_0+a_1+\cdots +a_{99}{)}^2}{a_1+a_2+\cdots +a_{100}}\ge a_0+a_2+\cdots +a_{99}=\sum _{n=1}^{100}{a}_{n-1}.$$ This yields that $$\sum _{n=1}^{100}\left(\frac{a_n}{a_{n-1}}-a_{n-1}\right)\le 50.$$ The rule given in the problem implies that $b_n=\frac{a_n}{a_{n-1}}-a_{n-1}$ for all $n=1,2,...,100$. Hence $S\le 50$. The equality holds when $a_i=1/2,i=0,1,...,100,b_i=1/2,i=1,2,...,100$.