If $a,b,c$ are the sides of a triangle, then $\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}$ is: If $a,b,c$ are the sides of a triangle, then $\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}$ is:
$A)$ $\le3$ , $B)$ $\ge3$, $(C)$ $\ge2$, $(D)$ $\le2$
My attempt is as follows:-
$$\dfrac{1}{2}\left(\dfrac{a}{s-a}+\dfrac{b}{s-b}+\dfrac{c}{s-c}\right)$$
Let $y=\dfrac{a}{s-a}+\dfrac{b}{s-b}+\dfrac{c}{s-c}$
$$A.M\ge H.M$$
$$\dfrac{\dfrac{s-a}{a}+\dfrac{s-b}{b}+\dfrac{s-c}{c}}{3}\ge \dfrac{3}{y}$$
$$\dfrac{\dfrac{s\cdot (ab+bc+ca)}{abc}-3}{3}\ge\dfrac{3}{y}$$
$$y\ge\dfrac{9}{\dfrac{(a+b+c)(ab+bc+ca)}{2abc}-3}\tag{1}$$
Let $z=\dfrac{(a+b+c)(ab+bc+ca)}{abc}$
Equation $(1)$ will give us $y_{min}$, so for that we need to find maximum value of $z$
But unfortunately I was able to find minimum value of $z$ in the following way
$$\dfrac{a+b+c}{3}\ge \dfrac{3abc}{ab+bc+ca}$$
$$\dfrac{(a+b+c)(ab+bc+ca)}{abc}\ge 9$$
But nevertheless, I tried plugging this minimum value of $z$ into equation $(1)$ and I got $y\ge 6$ and as the original expression was $\dfrac{y}{2}$, so $\dfrac{y}{2}\ge 3$ and surprisingly this answer is correct. What am I missing here?

The answer is B). See my solution.
By your work we need to prove that:
$$\dfrac{9}{\dfrac{(a+b+c)(ab+bc+ca)}{2abc}-3}\tag{1}\geq6,$$ which is impossible because it's wrong (you got a reversed inequality).
Another way.
$$\sum_{cyc}\frac{a}{b+c-a}-3=\sum_{cyc}\left(\frac{a}{b+c-a}-1\right)=\sum_{cyc}\frac{a-b-(c-a)}{b+c-a}=$$
$$=\sum_{cyc}(a-b)\left(\frac{1}{b+c-a}-\frac{1}{c+a-b}\right)=\sum_{cyc}\frac{2(a-b)^2}{(b+c-a)(c+a-b)}\geq0.$$
