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Monday, January 12, 2015

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




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