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 inecuality:
Use Titu's Lemma and $a^2+b^2+c^2\ge ab+bc+ca$
$$\frac a{b+c}+\frac b{c+a}+\frac c{a+b}=\frac {a^2}{ab+ac}+\frac {b^2}{bc+ba}+\frac {c^2}{ca+bc}\ge$$
$$\ge\frac{(a+b+c)^2}{2(ab+bc+ca)}=\frac{a^2+b^2+c^2}{2(ab+bc+ca)}+1\ge\frac12+1=\frac32$$
The left inecuality is Nesbitt's inequality
