2015/005) if $l \ge m \ge n \gt 0$ show that

$\dfrac{l^2-m^2}{n}+ \dfrac{n^2-m^2}{l}+\dfrac{l^2-n^2}{m} \ge 4l - 3m +n$


as $l \ge m \ge n \gt 0$

we get $(l +m) \ge 2n$

or $\dfrac{l+m}{n} \ge 2$
or $\dfrac{l^2-m^2}{n} \ge 2(l-m) \cdots (1) $

further
 $(m+n ) \le  2l$
or $\dfrac{m+n }{l} \le 2$
or $\dfrac{m^2-n^2}{l} \le 2(m-n) $
or $\dfrac{n^2-m^2}{l} \ge 2(n-m) \cdots (2) $

also
$(l+n ) \ge  m$
or $\dfrac{l+n }{m} \ge 1$
or $\dfrac{l^2-n^2}{m} \ge l-n \cdots (3)$

adding (1), (2), (3) we get
$\dfrac{l^2-m^2}{n}+ \dfrac{n^2-m^2}{l}+\dfrac{l^2-n^2}{m} \ge 4l - 3m +n$

Note: 
This problem I have taken from http://mathhelpboards.com/challenge-questions-puzzles-28/inequality-challenge-x-12071.html#post65756  where I was the only person who answered the same