onlylove_math

Let$a;b;c \ge\ \dfrac {1}{\sqrt {3} }$ and $abc = 1$. Prove that:
$\dfrac {c(a + b)}{c^2 + (a + b)^2} + \dfrac {a(b + c)}{a^2 + (b + c)^2} + \dfrac {b(c + a)}{b^2 + (c + a)^2} \le\ \dfrac {1}{5}(a^3 + b^3 + c^3 + 3)$