IMC 2010 · Day 2 · P7

Apply induction on $n$. Considering the empty product as 1, we have equality for $n = 0$. Now assume that the statement is true for some $n$ and prove it for $n+1$. For $n+1$, the statement can be written as the sum of the inequalities $$1 + \frac{1}{a_0} \left( 1 + \frac{1}{a_1 - a_0} \right) \cdots \left( 1 + \frac{1}{a_n - a_0} \right) \le \left( 1 + \frac{1}{a_0} \right) \cdots \left( 1 + \frac{1}{a_n} \right)$$ (which is the induction hypothesis) and $$\tag{1} \frac{1}{a_0} \left( 1 + \frac{1}{a_1 - a_0} \right) \cdots \left( 1 + \frac{1}{a_n - a_0} \right) \cdot \frac{1}{a_{n+1} - a_0} \le \left( 1 + \frac{1}{a_0} \right) \cdots \left( 1 + \frac{1}{a_n} \right) \cdot \frac{1}{a_{n+1}}.$$ Hence, to complete the solution it is sufficient to prove (1).

To prove (1), apply a second induction. For $n = 0$, we have to verify $$\frac{1}{a_0} \cdot \frac{1}{a_1 - a_0} \le \left( 1 + \frac{1}{a_0} \right) \frac{1}{a_1}.$$ Multiplying by $a_0 a_1 (a_1 - a_0)$, this is equivalent with $$\begin{align*} a_1 &\le (a_0 + 1)(a_1 - a_0) \\ a_0 &\le a_0 a_1 - a_0^2 \\ 1 &\le a_1 - a_0. \end{align*}$$ For the induction step it is sufficient that $$\left( 1 + \frac{1}{a_{n+1} - a_0} \right) \cdot \frac{a_{n+1} - a_0}{a_{n+2} - a_0} \le \left( 1 + \frac{1}{a_{n+1}} \right) \cdot \frac{a_{n+1}}{a_{n+2}}.$$ Multiplying by $(a_{n+2} - a_0) a_{n+2}$, $$\begin{align*} (a_{n+1} - a_0 + 1) a_{n+2} &\le (a_{n+1} + 1)(a_{n+2} - a_0) \\ a_0 &\le a_0 a_{n+2} - a_0 a_{n+1} \\ 1 &\le a_{n+2} - a_{n+1}. \end{align*}$$

Remark 1. It is easy to check from the solution that equality holds if and only if $a_{k+1} - a_k = 1$ for all $k$.

Remark 2. The statement of the problem is a direct corollary of the identity $$1 + \sum_{i=0}^{n} \frac{1}{x_i} \prod_{j \ne i} \left( 1 + \frac{1}{x_j - x_i} \right) = \prod_{i=0}^{n} \left( 1 + \frac{1}{x_i} \right).$$