Inequality. $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2}$ prove the following inequality: 
$$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2},$$ for $a,b,c$ real positive numbers. 
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Let $$M=\dfrac{a^3}{a+b}+\dfrac{b^3}{b+c}+\dfrac{c^3}{c+a}$$ be the original expression, we introduce its "conjugate" 
$$N=\dfrac{ab^2}{a+b}+\dfrac{bc^2}{b+c}+\dfrac{ca^2}{c+a}$$
Direct computing, and using $a^2+b^2+c^2\geq ab+bc+ca$, yields
$$M-N =\frac{a(a^2-b^2)}{a+b}+\frac{b(b^2-c^2)}{b+c}+\frac{c(c^2-a^2)}{c+a}\\=a(a-b)+b(b-c)+c(c-a)\\= a^2+b^2+c^2-ab-bc-ca\geq 0$$
Therefore $M\geq N$. Also using $2(x^2+y^2)\geq (x+y)^2$
$$M+N=\frac{a(a^2+b^2)}{a+b}+\frac{b(b^2+c^2)}{b+c}+\frac{c(c^2+a^2)}{c+a}\\ \geq \frac{a(a+b)+b(b+c)+c(c+a)}{2} \\ =\frac{a^2+b^2+c^2+ab+bc+ca}{2} \\ \geq ab+bc+ca$$
Hence we have both $M\geq N$ and $M+N\geq ab+bc+ca$. It follows that $M\geq \dfrac{ab+bc+ca}{2}$.
