Chinese Mathematical Olympiad

(1) Set $a_1=a$ and $b_1=b$ ($a,b>0$). The recurrence formula of $a_n$ implies that, for any positive integer $n\ge 2$, we have $$\frac{1}{a_n-a_{n+1}}=1+\sum _{i=1}^n\frac{1}{a_i}⟹\frac{1}{a_n-a_{n+1}}=\frac{1}{a_{n-1}-a_n}+\frac{1}{a_n}.$$ That is, $\frac{a_n}{a_n-a_{n-1}}=\frac{a_n}{a_{n-1}-a_n}+1=\frac{a_{n-1}}{a_{n-1}-a_n}$. From this, we obtain $\frac{a_{n+1}}{a_n}=\frac{a_n}{a_{n-1}}$. So $\{a_n\}$ is a geometric sequence with common ratio $\frac{a}{a+1}$, and the general term is $$a_n=\frac{a^n}{(a+1{)}^{n-1}}.$$ Similarly, the recurrence formula for $b_n$ implies that $$\frac{1}{b_{n+1}-b_n}=1+\sum _{i=1}^n\frac{1}{b_i}=\frac{1}{b_n-b_{n-1}}+\frac{1}{b_n}.$$ Thus, $\frac{b_n}{b_{n+1}-b_n}=\frac{b_n}{b_{n-1}-b_n}+1$. By induction, we have $\frac{b_n}{b_{n+1}-b_n}=\frac{b_1}{b_2-b_1}+2(n-1)=b+2n-1$. Therefore, $\frac{b_{n+1}}{b_n}=\frac{b+2n}{b+2n-1}$. From this, we get $b_n=b{\prod }_{k=1}^{n-1}\frac{b+2k}{b+2k-1}$. The condition $a_{100}{b}_{100}=a_{101}{b}_{101}$ in (1) implies that $\frac{a_{100}}{a_{101}}=\frac{b_{101}}{b_{100}}$, i.e. $\frac{a+1}{a}=\frac{b+200}{b+199}$. So $a=b+199$ and thus $a_1-b_1=a-b=199$.

(2) Method 1. It is clear that $\{a_n\}$ is monotonically decreasing and $\{b_n\}$ is monotonically increasing. Together with the condition $a_{100}=b_{99}$, we deduce that () $a_1>a_2>\cdots >a_{99}>a_{100}=b_{99}>b_{98}>\cdots >b_1>0$. In turn, we deduce that $$a_{99}=a_{100}+\frac{1}{1+{\sum }_{i=1}^{99}\frac{1}{a_i}}>b_{99}+\frac{1}{1+{\sum }_{i=1}^{99}\frac{1}{b_i}}=b_{100}.$$

Method 2. Since $a_{100}=b_{99}$, we have $$(\ast \ast )\frac{a^{100}}{(a+1{)}^{99}}=\frac{b(b+2)\cdots (b+196)}{(b+1)(b+3)\cdots (b+195)}.$$ To prove $a_{100}+b_{100}>a_{101}+b_{101}$, it is equivalent to prove that $$\begin{array}{rl}{a}_{100}-a_{101}& >b_{101}-b_{100}\\ ⟺\frac{a_{100}-a_{101}}{a_{100}}& >\frac{b_{101}-b_{100}}{b_{99}}\\ ⟺1-\frac{a}{a+1}& >\frac{(b+198)(b+200)}{(b+197)(b+199)}-\frac{b+198}{b+197}\\ ⟺\frac{1}{a+1}& >\frac{b+198}{(b+197)(b+199)}\\ (\ast )& ⟺a<b+197-\frac{1}{b+198}.\end{array}$$ We prove this by contradiction. Suppose that $a\ge b+197-\frac{1}{b+198}$, in particular, $a>b+196$. We have $$\begin{array}{rl}\frac{a^2}{a+1}-(b+196)& =\frac{a(a-b-196)-(b+196)}{a+1}\\ & \ge \frac{1}{a+1}\left(\left(b+197-\frac{1}{b+198}\right)\left(1-\frac{1}{b+198}\right)-(b+196)\right)\\ & >\frac{1}{a+1}\left(\left(b+197-\frac{1}{b+197}\right)\frac{b+197}{b+198}-(b+196)\right)=0.\end{array}$$ But note that $\frac{a}{a+1}>\frac{b+194}{b+195}>\frac{b+192}{b+193}>\cdots >\frac{b}{b+1}$, we have $$\frac{a^{100}}{(a+1{)}^{99}}={\left(\frac{a}{a+1}\right)}^{98}\frac{a^2}{a+1}>\frac{b}{b+1}\cdots \frac{b+194}{b+195}(b+196),$$ contradicting with ()! So () holds, and thus $a_{100}+b_{100}>a_{101}+b_{101}$.