, 2015

Using the condition $a + b + c = 1$, we have
$$\begin{aligned} \left(\frac{a b}{a + b} + \frac{b c}{b + c} + \frac{c a}{c + a}\right) &= (a + b + c)\left(\frac{a b}{a + b} + \frac{b c}{b + c} + \frac{c a}{c + a}\right) \\ &= (a b + b c + c a) + a b c\left(\frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a}\right) \\ &\geq (a b + b c + c a) + \frac{9 a b c}{2(a + b + c)} \\ &\geq (a b + b c + c a) + \frac{9}{2} a b c, \end{aligned}$$
by Cauchy-Schwarz inequality. Therefore, it remains to prove that
$$9 a b c + 1 \geq 4(a b + b c + c a).$$
Using again the condition $a + b + c = 1$, the last inequality becomes
$$9 a b c + (a + b + c)^2 \geq 4(a + b + c)(a b + b c + c a),$$
which is equivalent, by simple algebraic manipulations, to
$$9 a b c + a^2 + b^2 + c^2 \geq 2(a b + b c + c a)$$
Using a third time the condition $a + b + c = 1$, the obtained inequality becomes
$$9 a b c + (a + b + c)(a^2 + b^2 + c^2) \geq 2(a + b + c)(a b + b c + c a)$$
which is equivalent, by simple algebraic manipulations, to
$$3 a b c + a^3 + b^3 + c^3 \geq a^2(b + c) + b^2(c + a) + c^2(a + b).$$
This last inequality is nothing but Schur's inequality. This solves the problem.