TITLE: $a,b,c \geq 0$ and $a+b+c=3$ prove that $\frac{a^2+bc}{b+ac} + \frac{b^2+ac}{c+ab} + \frac{c^2+ab}{a+bc} \geq 3$
QUESTION [8 upvotes]: $a,b,c \geq 0$ and $a+b+c=3$ prove that $\frac{a^2+bc}{b+ac} + \frac{b^2+ac}{c+ab} + \frac{c^2+ab}{a+bc} \geq 3$
can anyone help me solve this problem,i've tried to use C-S and also AM-GM for couple in cycle.Sorry for my bad English that caused me problem to explain my trying.

REPLY [5 votes]: $$\frac{a^2+bc}{b+ac} + \frac{b^2+ac}{c+ab} + \frac{c^2+ab}{a+bc} \geq 3$$
$$\iff \frac{a^2+bc}{3b+3ac} + \frac{b^2+ac}{3c+3ab} + \frac{c^2+ab}{3a+3bc} \geq 1$$
$$\iff \frac{a^2+bc}{(a+b+c)b+3ac} + \frac{b^2+ac}{(a+b+c)c+3ab} + \frac{c^2+ab}{(a+b+c)a+3bc} \geq 1$$
$$\Leftarrow \frac{a^2+bc}{(a+b+c)b+(a^2+ac+c^2)} + \frac{b^2+ac}{(a+b+c)c+(a^2+ab+b^2)} + \frac{c^2+ab}{(a+b+c)a+(b^2+bc+c^2)} = 1$$