Prove that $\dfrac32\leq \dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} < 2.$ where $a$ ,$b$ ,$c$ are sides of Triangle
I tried using sine rule which makes up the the expression $\sum \dfrac{\sin{A}}{\sin{B}+\sin{C}}$. Now i dont how to find the range of $\sum \dfrac{\sin{A}}{\sin{B}+\sin{C}}$. Any ideas(hint) or solution would be really appreciated.
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Making $b = x a$ and $c=y a$ the conditions read
$$ \frac 32\le\frac{1}{x+y}+\frac{x}{1+y}+\frac{y}{1+x}\le 2 \ \ \text{s. t.}\ \ x + y\ge 1 $$
so making the lagrangian
$$ L(x,y,\lambda) = \frac{1}{x+y}+\frac{x}{1+y}+\frac{y}{1+x}+\lambda(x+y-1) $$
We have two stationary points. One at $x^*=1, y^* = 1$ corresponding to a minimum $(\frac 32)$ and $x^* = \frac 12, y^* = \frac 12$ corresponding to a maximum $\frac 53$ so we have
$$ \frac 32\le \frac 32\le \frac 53\le 2 $$
As pointed out in a comment, you can use Nesbitt's inequality (which is ultimately an application of the Arithmetic-Geometric mean inequality) for the lower bound.
For the upper bound, note that since $a < b+c$, we have $\frac{a}{b+c} < \frac{a+a}{b+c+a} = \frac{2a}{a+b+c}$. Now apply similar inequalities to the other terms and sum them.