Inequality based on triangle: $\frac{3}{2}\le\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$ 
If $a,b,c$ are sides of a triangle, prove that ${3\over2 }\le {{a\over b+c}} + {{b\over c+a}}+{{c\over a+b}} \lt 2$ .


The left inequality.
$\sum\limits_{cyc}\frac{a}{b+c}-\frac{3}{2}=\sum\limits_{cyc}\left(\frac{a}{b+c}-\frac{1}{2}\right)=\sum\limits_{cyc}\frac{a-b-(c-a)}{2(b+c)}=\sum\limits_{cyc}\frac{(a-b)^2}{2(a+c)(b+c)}\geq0$.
The right inequality.
$\sum\limits_{cyc}\frac{a}{b+c}<\sum\limits_{cyc}\frac{a+a}{a+b+c}=2$
