$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq \frac{3}{2}$

Let $a,b,c$ be positive real numbers. Prove the inequality $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq \frac{3}{2}.$$

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From $(a-b)^2\geq 0$, we know $\dfrac{a}{b}+\dfrac{b}{a}\geq 2$. Equality occurs iff $a=b$. Using this inequality, we then have $$\frac{a+b}{b+c}+\frac{b+c}{a+b}+\frac{b+c}{c+a}+\frac{c+a}{b+c}+\frac{c+a}{a+b}+\frac{a+b}{c+a}\geq 6.$$ We rewrite the inequality as $$\bigg(\frac{a+b}{b+c}+\frac{c+a}{b+c}\bigg)+\bigg(\frac{a+b}{c+a}+\frac{b+c}{c+a}\bigg)+\bigg(\frac{b+c}{a+b}+\frac{c+a}{a+b}\bigg) \geq 6,$$ which means $$\frac{2a}{b+c}+1+\frac{2b}{c+a}+1+\frac{2c}{a+b}+1\geq 6,$$ or $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq \frac{3}{2}$$ as required.

Equality occurs iff $\dfrac{a+b}{b+c}=\dfrac{b+c}{a+b},\dfrac{b+c}{c+a}=\dfrac{c+a}{b+c},\dfrac{c+a}{a+b}=\dfrac{a+b}{c+a}$ from where we deduce $a=b=c$.

Takeaways:
1. Make use of known inequalities.
2. Reformulation of the problem may give us more insight: rewrite the inequality as $$\frac{2a}{b+c}+\frac{2b}{a+c}+\frac{2c}{a+b}\geq 3.$$ 3. We can start with $\dfrac{a+b}{b+c}+\dfrac{b+c}{a+b}\geq 2$. Since we want $\dfrac{2a}{b+c}$, we add $\dfrac{b+c}{c+a}+\dfrac{c+a}{b+c}$ to $\dfrac{a+b}{b+c}+\dfrac{b+c}{a+b}$. Note also the symmetry of the problem. We should have $2\cdot 3=6$ terms.
4. Don't forget the equality case.