# Proof by induction for tricky double summation: $(\sum_{k=1}^n x_k)\cdot(\sum_{k=1}^n \frac{1}{x_k})\ge{n^2}$ [duplicate]

let $x_1....x_n$ be positive integers. Prove by induction the following for natural numbers n:

$(\sum_{k=1}^n x_k)\cdot(\sum_{k=1}^n \frac{1}{x_k})\ge{n^2}$

Hint: for all positive integers a,b: $\frac{a}{b}+\frac{b}{a}\ge2$

• I highly suspect that OP is supposed to prove that given expression is $\geq n^2$. – Wojowu Oct 31 '15 at 14:47
• Oh I'm sorry I missed a part out! I shall fix now – Cam Mack Oct 31 '15 at 15:35
• @JackD'Aurizio I have fixed it now – Cam Mack Oct 31 '15 at 15:37
• @Wojowu yes you are right – Cam Mack Oct 31 '15 at 15:38

$$\left(\sum_{i=1}^n x_i\right)\left(\sum_{i=1}^n \frac{1}{x_i}\right)$$ $$=\left(\sum_{i=1}^{n-1} x_i + x_n\right)\left(\sum_{i=1}^{n-1} \frac{1}{x_i} + \frac{1}{x_n}\right)$$ $$=\left(\sum_{i=1}^{n-1} x_i\right)\left(\sum_{i=1}^{n-1} \frac{1}{x_i}\right) + \frac{1}{x_n}\left(\sum_{i=1}^{n-1} x_i\right) + x_n\left(\sum_{i=1}^{n-1}\frac{1}{x_i}\right) + 1$$ Now $$\left(\sum_{i=1}^{n-1} x_i\right)\left(\sum_{i=1}^{n-1} \frac{1}{x_i}\right) \geq (n-1)^2$$ and
$$\frac{1}{x_n}\left(\sum_{i=1}^{n-1} x_i\right) + x_n\left(\sum_{i=1}^{n-1} \frac{1}{x_i}\right) = \sum_{i=1}^{n-1}\left(\frac{x_i}{x_n} + \frac{x_n}{x_i}\right) \geq 2(n-1)$$ so that $$\left(\sum_{i=1}^{n-1} x_i\right)\left(\sum_{i=1}^{n-1} \frac{1}{x_i}\right) + \frac{1}{x_n}\left(\sum_{i=1}^{n-1} x_i\right) + x_n\left(\sum_{i=1}^{n-1}\frac{1}{x_i}\right) + 1 \geq (n-1)^2 + 2(n-1) + 1 = ((n-1) + 1)^2 = n^2$$