An inequality with a product of two sums

**tttcomrader** · January 25th 2011, 11:18 AM

I'm looking a problem that states:

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

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

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

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

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

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

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

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

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

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

Any hints? Thank you very much!

**Jose27** · January 25th 2011, 03:42 PM

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 $a=(\sqrt{x_1},...,\sqrt{x_n}),\ b=(\frac{1}{\sqrt{x_1}},...,\frac{1}{\sqrt{x_n}})$ where you give $\mathbb{R}^n$ the usual interior product and norm.

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