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: $\frac{a}{b}+\frac{b}{a}\ge2$
$\begingroup$
$\endgroup$
4
-
1$\begingroup$ I highly suspect that OP is supposed to prove that given expression is $\geq n^2$. $\endgroup$– WojowuOct 31, 2015 at 14:47
-
$\begingroup$ Oh I'm sorry I missed a part out! I shall fix now $\endgroup$– Cam MackOct 31, 2015 at 15:35
-
$\begingroup$ @JackD'Aurizio I have fixed it now $\endgroup$– Cam MackOct 31, 2015 at 15:37
-
$\begingroup$ @Wojowu yes you are right $\endgroup$– Cam MackOct 31, 2015 at 15:38
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
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 $$\left(\sum_{i=1}^{n-1} x_i\right)\left(\sum_{i=1}^{n-1} \frac{1}{x_i}\right) + \frac{1}{x_n}\left(\sum_{i=1}^{n-1} x_i\right) + x_n\left(\sum_{i=1}^{n-1}\frac{1}{x_i}\right) + 1 \geq (n-1)^2 + 2(n-1) + 1 = ((n-1) + 1)^2 = n^2$$
-
$\begingroup$ I will mention that a similar proof can be found in this answer: Proof that $\left(\sum^n_{k=1}x_k\right)\left(\sum^n_{k=1}y_k\right)\geq n^2$. $\endgroup$ Jan 21, 2019 at 14:34