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
 A: Your solution is right.
Also, SOS helps:
$$\sum_{cyc}\frac{a}{b+c}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{b+c}-\frac{1}{2}\right)=\sum_{cyc}\frac{2a-b-c}{2(b+c)}=$$
$$=\sum_{cyc}\frac{a-b-(c-a)}{2(b+c)}=\sum_{cyc}\left(\frac{a-b}{2(b+c)}-\frac{c-a}{2(b+c)}\right)=$$
$$=\sum_{cyc}\left(\frac{a-b}{2(b+c)}-\frac{a-b}{2(c+a)}\right)=\sum_{cyc}(a-b)\left(\frac{1}{2(b+c)}-\frac{1}{2(c+a)}\right)=$$
$$=\sum_{cyc}\frac{(a-b)^2}{2(a+c)(b+c)}\geq0.$$
Now we see that the starting inequality is true for any reals $a$, $b$ and $c$ such that $ab+ac+bc>0.$
Also, there is a solution by AM-GM, by C-S, by TL, by $uvw$ and by more and more and more.
A: Solution by Buffalo Way method.
Let $a=\min\{a,b,c\},$ $b=a+u$ and $c=a+v$.
Thus, $$2\prod_{cyc}(a+b)\left(\sum_{cyc}\frac{a}{b+c}-\frac{3}{2}\right)=4(u^2-uv+v^2)a+(u+v)(2u^2-3uv+2v^2)\geq0.$$
A: Solution by the Tangent Line method.
Since our inequality is homogeneous, we can assume that $a+b+c=3$ and we onbtain: $$\sum_{cyc}\frac{a}{b+c}-\frac{3}{2}=\sum_{cyc}\frac{a}{3-a}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{3-a}-\frac{1}{2}\right)=$$
$$\sum_{cyc}\frac{3(a-1)}{2(3-a)}=\frac{3}{2}\sum_{cyc}\left(\frac{a-1}{3-a}-\frac{1}{2}(a-1)\right)=\frac{9}{4}\sum_{cyc}\frac{(a-1)^2}{3-a}\geq0.$$
A: Another proof$:$
Due to homogeneous, assume $a+b+c=1.$
Let $p=a+b+c=1,q=\dfrac{1-t^2}{3} \quad(\, t\in [\,0,1\,]\,),r=abc.$
Need to prove$:$ $$\frac73\,{t}^{2}+9\,r-\frac13 \geqslant 0$$
Since $$r\geqslant \dfrac{1}{27} \left( 1-2t \right) \left( 1+t \right) ^{2}$$
We need to prove$:$ $$\dfrac{2}{3} t^2(2-t) \geqslant 0,$$
which is true since $t \in [\,0,\,1\,].$
See also here.
There is also a proof by SS (SOS - Schur) method.
$$\text{LHS}-\text{RHS}={\frac {2\, \left( a-b \right) ^{2} \left( a+b \right) + \left( a
-c \right)  \left( b-c \right)  \left( a+b+2\,c \right) }{2 \left( b+c
 \right)  \left( c+a \right)  \left( a+b \right) }} \geqslant 0,$$
which is obvious if $c\equiv \min\{a,b,c\}.$
