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: $\frac{a}{b}+\frac{b}{a}\ge2$
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Sign up to join this communitylet $x_1....x_n$ 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: $\frac{a}{b}+\frac{b}{a}\ge2$
The inductive step should be
$$\left(\sum_{i=1}^n x_i\right)\left(\sum_{i=1}^n \frac{1}{x_i}\right)$$ $$=\left(\sum_{i=1}^{n-1} x_i + x_n\right)\left(\sum_{i=1}^{n-1} \frac{1}{x_i} + \frac{1}{x_n}\right)$$ $$=\left(\sum_{i=1}^{n-1} x_i\right)\left(\sum_{i=1}^{n-1} \frac{1}{x_i}\right) + \frac{1}{x_n}\left(\sum_{i=1}^{n-1} x_i\right) + x_n\left(\sum_{i=1}^{n-1}\frac{1}{x_i}\right) + 1$$ 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 $$\left(\sum_{i=1}^{n-1} x_i\right)\left(\sum_{i=1}^{n-1} \frac{1}{x_i}\right) + \frac{1}{x_n}\left(\sum_{i=1}^{n-1} x_i\right) + x_n\left(\sum_{i=1}^{n-1}\frac{1}{x_i}\right) + 1 \geq (n-1)^2 + 2(n-1) + 1 = ((n-1) + 1)^2 = n^2$$