<s>[INST] 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
 [/INST] 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.
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