WakeUp wrote: Let $a,b$ and $c$ be the side lengths of a triangle. Prove that: \[\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}<\frac{1}{16}\] After substitutions $a=y+z$, $b=x+z$ and $c=x+y$ for positives $x$, $y$ and $z$ and after expanding we obtain an obvious inequality.

Can anyone show me this known inequality? After expanding I got $15(ac^2 + a^2b + b^2c) \le 17(bc^2+ ab^2 + a^2c) + 2abc$

see here http://www.artofproblemsolving.com/Forum/blog.php?u=40112&b=65576

arqady wrote: After substitutions $a=y+z$, $b=x+z$ and $c=x+y$ for positives $x$, $y$ and $z$ and after expanding we obtain an obvious inequality. $\frac{y-x}{x+y+2z}+\frac{z-y}{2x+y+z}+\frac{x-z}{x+2y+z}<\frac{y-x}{x+y+z}+\frac{z-y}{x+y+z}+\frac{x-z}{x+y+z}=0$

WolfusA wrote: arqady wrote: After substitutions $a=y+z$, $b=x+z$ and $c=x+y$ for positives $x$, $y$ and $z$ and after expanding we obtain an obvious inequality. $\frac{y-x}{x+y+2z}+\frac{z-y}{2x+y+z}+\frac{x-z}{x+2y+z}<\frac{y-x}{x+y+z}+\frac{z-y}{x+y+z}+\frac{x-z}{x+y+z}=0$ The first step will be true only if $y-x, z-y, x-z$ all are positive which is obviously false.

My mistake.