fattypiggy123 wrote: Let $a,b,c > 0$, prove that $\frac{(a-b)^2}{(c+a)(c+b)} + \frac{(b-c)^2}{(a+b)(a+c)} + \frac{(c-a)^2}{(b+c)(b+a)} \geq \frac{(a-b)^2}{a^2+b^2+c^2} $ Just do Cauchy!

$ (\sum{|b-c|})^{2}\geq4(a-b)^{2} $ $ \sum{(a+b)(b+c)}=\sum{a^{2}}+3\sum{bc}\leq4\sum{a^{2}} $ $ LHS\geq\frac{(\sum{|b-c|})^{2}}{\sum{(a+b)(b+c)}} $ $ \geq\frac{(a-b)^{2}}{\sum{a^{2}}} $

In form, I think If $a,b,c,$ are a triangle's side lengths , prove that $\frac{ |c^2-a^2 |}{b} + \frac{|a^2-b^2 |}{c} \geq \frac{|b^2-c^2 |}{a} $

sqing wrote: In form, I think If $a,b,c,$ are a triangle's side lengths , prove that $\frac{ |c^2-a^2 |}{b} + \frac{|a^2-b^2 |}{c} \geq \frac{|b^2-c^2 |}{a} $ We just assume $ b\geq a\geq c $ and $ \frac{a^{2}-b^{2}}{c}=c-2a\cos{B} $ and $ b=a\cos{C}+c\cos{A} $ it's enough!

It was my mistake

Pantum wrote: sqing wrote: In form, I think If $a,b,c,$ are a triangle's side lengths , prove that $\frac{ |c^2-a^2 |}{b} + \frac{|a^2-b^2 |}{c} \geq \frac{|b^2-c^2 |}{a} $ We just assume $ b\geq a\geq c $ and $ \frac{a^{2}-b^{2}}{c}=c-2a\cos{B} $ and $ b=a\cos{C}+c\cos{A} $ it's enough! thank you! We just assume $ b\geq a\geq c $，why?

\[ \frac{(a-b)^{2}}{(c+a)(c+b)}+\frac{(b-c)^{2}}{(a+b)(a+c)}+\frac{(c-a)^{2}}{(b+c)(b+a)}= \] \[=\frac{(a-b)^{2}}{(c+a)(c+b)}+\frac{(c-b)^{2}}{(a+b)(a+c)}+\frac{(a-c)^{2}}{(b+c)(b+a)}\ge\] \[\ge \frac{4(a-b)^2}{\sum{(a+b)(a+c)}}\ge \frac{(a-b)^2}{\sum{a^2}}\]

sqing wrote: We just assume $ b\geq a\geq c $，why? Because the inequality would were strongest when $ b\geq a\geq c $

Actually wouldn't it be when $a \ge c \ge b$?

Markomak1 wrote: Actually wouldn't it be when $a \ge c \ge b$? Actually,we can prove it simply without assum. $ \frac{|a^{2}-b^{2}|}{c}=|b\cos{A}-a\cos{B}|=\frac{|\sin{(A-B)|}}{2R} $ so we just prove $ \sin{|A-B|}+\sin{|C-A|}\geq\sin{|B-C|} $ easily to prove!

I am sorry pantum, i actually got confused on which post you were commenting, on the problem you proposed with triangle sides, the case you suggested is the strongest indeed.