How to prove $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} <2$? Prove the inequality for a triangle with sides $a,b,c$ we have $$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} <2$$
Trial: Since $a,b,c$ are sides of a triangle I know  $a+b>c,b+c>a,a+c>b$  

That $a,b,c$ are sides of a triangle is equivalent to $a=x+y$, $b=y+z$ and $c=z+x$. Therefore we can rewrite our inequality as follows:
$$
\dfrac{x+y}{x+y+2z}+\dfrac{y+z}{2x+y+z}+\dfrac{z+x}{x+2y+z} <2
$$
Now, we have $\dfrac{x+y}{x+y+2z}<\dfrac{x+y}{x+y+z}$, $\dfrac{y+z}{2x+y+z}<\dfrac{y+z}{x+y+z}$ and $\dfrac{z+x}{x+2y+z}<\dfrac{z+x}{x+y+z}$ and therefore:
$$
\dfrac{x+y}{x+y+2z}+\dfrac{y+z}{2x+y+z}+\dfrac{z+x}{x+2y+z} <\dfrac{x+y}{x+y+z}+\dfrac{y+z}{x+y+z}+\dfrac{z+x}{x+y+z} =2
$$
