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 ${3\over2 }\le {{a\over b+c}} + {{b\over c+a}}+{{c\over a+b}} \lt 2$ .


We show the right side of the inequality, that is,
\begin{align*}
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} < 2, 
\end{align*}
which is equivalent to 
\begin{align*}
\frac{b+c-a}{b+c}+\frac{c+a-b}{c+a}+\frac{a+b-c}{a+b} > 1.
\end{align*}
Let $x=b+c-a$, $y=c+a-b$, and $z=a+b-c$. Then $x>0$, $y>0$, and $z>0$. Moreover,
\begin{align*}
\frac{b+c-a}{b+c}+\frac{c+a-b}{c+a}+\frac{a+b-c}{a+b} &= \frac{x}{x+\frac{y+z}{2}}+\frac{y}{y+\frac{x+z}{2}}+\frac{z}{z+\frac{x+y}{2}}\\
&=\frac{x^2}{x^2+x\frac{y+z}{2}}+\frac{y^2}{y^2+y\frac{x+z}{2}}+\frac{z^2}{z^2+z\frac{x+y}{2}}\\
&\ge\frac{(x+y+z)^2}{x^2+y^2+z^2+xy+xz+yz} > 1.
\end{align*}
We are done.
