Induction on $ n$ $ n=2$ $ \frac {1}{2} + \frac {1}{3} + \frac {1}{2 \cdot 3} = 1$ Suppose that you have a solution $ x_i$ for $ n$ $ \frac {1}{x_{1}} + \frac {1}{x_{2}} + \cdots + \frac {1}{x_{n}} + \frac {1}{x_{1}x_{2}\cdots x_{n}} = 1$ Defining $ y_i=x_i$ for $ i=1,2, \cdots n$ $ y_{n+1}= x_{1}x_{2}\cdots x_{n}+1$ $ \frac {1}{y_{1}} + \frac {1}{y_{2}} + \cdots + \frac {1}{y_{n}}+ \frac {1}{y_{n+1}} + \frac {1}{y_{1}y_{2}\cdots y_{n+1}}=$ $ \frac {1}{x_{1}} + \frac {1}{x_{2}} + \cdots + \frac {1}{x_{n}}+ \frac {1}{x_{1}x_{2}\cdots x_{n}+1} + \frac {1}{x_{1}x_{2}\cdots x_{n} (x_{1}x_{2}\cdots x_{n}+1)}=$ $ 1- \frac {1}{x_{1}x_{2}\cdots x_{n}} + \frac {1}{x_{1}x_{2}\cdots x_{n}+1} + \frac {1}{x_{1}x_{2}\cdots x_{n} (x_{1}x_{2}\cdots x_{n}+1)}=$ $ 1+\frac{1+x_{1}x_{2}\cdots x_{n}-(x_{1}x_{2}\cdots x_{n}+1)}{x_{1}x_{2}\cdots x_{n} (x_{1}x_{2}\cdots x_{n}+1)}=1$

Great problem, and also great solution. Where does the idea of defining $y_i = x_i$ for $i=1,2,...,n$ and $y_{n+1} = x_1 x_2 ... x_n + 1$ comes from?

silvergrasshopper wrote: Great problem, and also great solution. Where does the idea of defining $y_i = x_i$ for $i=1,2,...,n$ and $y_{n+1} = x_1 x_2 ... x_n + 1$ comes from? Thanks! The idea comes from trying to use induction from smaller cases, defining $y_i = x_i$ for $i=1,2,...,n$ and then getting the value of $y_{i+1}$

http://mathworld.wolfram.com/SylvestersSequence.html

mszew wrote: Induction on $ n$ $ n=2$ $ \frac {1}{2} + \frac {1}{3} + \frac {1}{2 \cdot 3} = 1$ Suppose that you have a solution $ x_i$ for $ n$ $ \frac {1}{x_{1}} + \frac {1}{x_{2}} + \cdots + \frac {1}{x_{n}} + \frac {1}{x_{1}x_{2}\cdots x_{n}} = 1$ Defining $ y_i=x_i$ for $ i=1,2, \cdots n$ $ y_{n+1}= x_{1}x_{2}\cdots x_{n}+1$ $ \frac {1}{y_{1}} + \frac {1}{y_{2}} + \cdots + \frac {1}{y_{n}}+ \frac {1}{y_{n+1}} + \frac {1}{y_{1}y_{2}\cdots y_{n+1}}=$ $ \frac {1}{x_{1}} + \frac {1}{x_{2}} + \cdots + \frac {1}{x_{n}}+ \frac {1}{x_{1}x_{2}\cdots x_{n}+1} + \frac {1}{x_{1}x_{2}\cdots x_{n} (x_{1}x_{2}\cdots x_{n}+1)}=$ $ 1- \frac {1}{x_{1}x_{2}\cdots x_{n}} + \frac {1}{x_{1}x_{2}\cdots x_{n}+1} + \frac {1}{x_{1}x_{2}\cdots x_{n} (x_{1}x_{2}\cdots x_{n}+1)}=$ $ 1+\frac{1+x_{1}x_{2}\cdots x_{n}-(x_{1}x_{2}\cdots x_{n}+1)}{x_{1}x_{2}\cdots x_{n} (x_{1}x_{2}\cdots x_{n}+1)}=1$ where $ y_{n+1}= x_{1}x_{2}\cdots x_{n}+1$