Traingle inequalty

By Ravi's substitution, let $a=x+y,b=y+z,c=z+x$ .

So, $\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}$

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

$<\dfrac{x+y}{x+y+z}+\dfrac{y+z}{y+z+x}+\dfrac{z+x}{z+x+y}$

$=\dfrac{2(x+y+z)}{x+y+z}=2$

Let $S = a+b+c =$ perimeter, It is known that the sum of any two sides of a triangle $> \dfrac{S}{2} \ \$ . Then $\begin{aligned} 2(a+b) &> S\\ 2(a+c) &> S\\ 2(b+c) &> S\\ \end{aligned} \implies \dfrac{a}{b+c} + \dfrac{b}{a+c} +\dfrac{c}{a+b} < \dfrac{2(a+b+c)}{(a+b+c)} = 2 \\$

If $a,b, c$ are sides of a triangle, then the minimum of

$\dfrac{a}{b+c} + \dfrac{b}{a+c} +\dfrac{c}{a+b} =1.5 \text{ which is less than} \ 2$

By Inequalities we know that $\dfrac{a}{b+c} + \dfrac{b}{a+c} +\dfrac{c}{a+b} \geq \dfrac{3}{2}= 1.5$

So it can be also said that it is lesser than 2.

It So it falls in the range , $2 > \dfrac{a}{b+c} + \dfrac{b}{a+c} +\dfrac{c}{a+b} \geq \dfrac{3}{2}$