Inequality Challenge X: Prove $\ge 3l-4m+n$

[anemone]

There are real numbers $l,\,m,\,n$ such that $l\ge m\ge n >0$.

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

 

[kaliprasad]

this is not answered since months. here is my solution

Spoiler

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

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

 

[anemone]

this is not answered since months. here is my solution

Spoiler

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

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

Well done, kaliprasad! I didn't realize I had one problem left unanswered...fortunately you helped me to finish the unfinished business here...thanks, my friend!