\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
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