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!