Multiple proofs of $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}2$
One answer using AM-GM.
From AM-GM, we obtain two inequalities:
- $a^{\frac{3}2}+b^{\frac{3}2}+b^{\frac{3}2}\geq3a^\frac{1}2b$
- $a^{\frac{3}2}+c^{\frac{3}2}+c^{\frac{3}2}\geq3a^\frac{1}2c$
Add them up and we get:
$$2(a^{\frac{3}2}+b^{\frac{3}2}+c^{\frac{3}2})\geq3a^\frac{1}2(b+c)\iff\frac{a}{b+c}\geq\frac{a^\frac{3}2}{a^{\frac{3}2}+b^{\frac{3}2}+c^{\frac{3}2}}$$
Therefore $$\sum_{cyc}\frac{a}{b+c}\geq\frac{3}2\sum_{cyc}\frac{a^\frac{3}2}{a^{\frac{3}2}+b^{\frac{3}2}+c^{\frac{3}2}}=\frac{3}2$$QED.
Note: This inequality is called Nesbitt's inequality, and in this link you can see a variety of approaches (as already commented by Dr.Mathva). This approach isn't included in it, so I think you'll probably like this as well.
Because $$\frac{a}{b+c} - \frac{8a-b-c}{4(a+b+c)} = \frac{(2a-b-c)^2}{4(b+c)(a+b+c)} \geqslant 0,$$ so $$\frac{a}{b+c} \geqslant \frac{8a-b-c}{4(a+b+c)}.$$ Therefore $$\sum \frac{a}{b+c} \geqslant \sum \frac{8a-b-c}{4(a+b+c)} = \frac 32.$$ Note. In addition, you can see 45th-proof-Nesbitt.pdf (Vietnamese)