Is this alternative proof of the inequality $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$ correct? Prove that for all positive real numbers:
$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq\dfrac{3}{2}$$
This is same as this question but a different approach is used there whereas I want to verify my approach to this problem.
My Approach:
$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\Big(\dfrac{a}{b+c}+1\Big)+\Big(\dfrac{b}{c+a}+1\Big)+\Big(\dfrac{c}{a+b}+1\Big)-3$$
$$=(a+b+c)\Big[\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\Big]-3$$
By AM-HM inequality:
$$\dfrac{3}{\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}}\leq\dfrac{2(a+b+c)}{3}\Rightarrow (a+b+c)\Big[\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\Big]\geq \dfrac{9}{2}$$
$$(a+b+c)\Big[\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\Big]-3\geq \dfrac{3}{2}$$
$\therefore \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq\dfrac{3}{2}\space \forall\ a,b,c\in \mathbb R$ and $a,b,c>0$
Please check this approach and provide suggestions. Also please provide alternative solutions if available.
THANKS

Solution by Buffalo Way method.
Let $a=\min\{a,b,c\},$ $b=a+u$ and $c=a+v$.
Thus, $$2\prod_{cyc}(a+b)\left(\sum_{cyc}\frac{a}{b+c}-\frac{3}{2}\right)=4(u^2-uv+v^2)a+(u+v)(2u^2-3uv+2v^2)\geq0.$$
