Thread: An inequality with a product of two sums

1. An inequality with a product of two sums

I'm looking a problem that states:

$\displaystyle \sum ^n _{i=1} x_i \sum ^n_{j=1} \frac {1}{x_j} \geq n^2$

Now if I write them down term by term I would have the following:

$\displaystyle x_1 ( \frac {1}{x_1} + \frac {1}{x_2} + ... + \frac {1}{x_n} ) + . . . + x_n ( \frac {1}{x_1} + \frac {1}{x_2} + ... + \frac {1}{x_n} )$

which equals to $\displaystyle \frac {x_1}{x_1} + \frac {x_1}{x_2} + . . . + \frac {x_1}{x_n} + . . . + \frac {x_n}{x_n}$

which is equals to $\displaystyle \frac {x_1}{x_1} + . . . + \frac {x_n}{x_n} + \frac {x_1}{x_2} + \frac {x_1}{x_3} + . . . + \frac {x_n}{x_{n-1}}$

$\displaystyle = n + \frac {x_1}{x_2} + \frac {x_1}{x_3} + . . . + \frac {x_n}{x_{n-1}}$

But how would I show that the above expression is bigger than $\displaystyle n^2$?

I tried to use the Holder's inequality by applying the square roof on both side, but then I have:

$\displaystyle ( \sum ^n _{i=1} x_i)^{ \frac {1}{2} } )( \sum ^n_{j=1} \frac {1}{x_j} )^{ \frac {1}{2} } \geq n$

But I'm still stuck as the p and q are the same in this case.

Any hints? Thank you very much!

2. You already have the answer, what's the problem with Hölder's inequality?

To put it in easier terms apply the Cauchy-Schwartz inequality to the vectors $\displaystyle a=(\sqrt{x_1},...,\sqrt{x_n}),\ b=(\frac{1}{\sqrt{x_1}},...,\frac{1}{\sqrt{x_n}})$ where you give $\displaystyle \mathbb{R}^n$ the usual interior product and norm.