How to prove $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} <2$?

Question

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 $a+b>c,b+c>a,a+c>b$

Answer 1

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: $$ \dfrac{x+y}{x+y+2z}+\dfrac{y+z}{2x+y+z}+\dfrac{z+x}{x+2y+z} <\dfrac{x+y}{x+y+z}+\dfrac{y+z}{x+y+z}+\dfrac{z+x}{x+y+z} =2 $$

Answer 2

\begin{align*}\frac a{b+c}+\frac b{c+a}+\frac c{a+b}<2&\iff 3-2<1-\frac a{b+c}+1-\frac b{c+a}+1-\frac c{a+b}\\&\iff 1<\frac {b+c-a}{b+c}+\frac {c+a-b}{c+a}+\frac {a+b-c}{a+b}\end{align*} Now $$\frac {b+c-a}{b+c}+\frac {c+a-b}{c+a}+\frac {a+b-c}{a+b}>\frac {b+c-a}{a+b+c}+\frac {c+a-b}{a+b+c}+\frac {a+b-c}{a+b+c}=1$$ You need the triangle inequality to ensure the fractions on the left side are positive, so you decrease them by increasing the denominator.