$a,b,c \geq 0$ 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.
Asked
Active
Viewed 245 times
8
-
2This doesn't look good. If $a=b=c=1/3$ then $a^2<b,\dots$ and the inequality doesn't hold. – Quang Hoang Aug 07 '14 at 17:19
-
my bad,i've editted – Aerrozard Aug 07 '14 at 17:23
1 Answers
5
$$\frac{a^2+bc}{b+ac} + \frac{b^2+ac}{c+ab} + \frac{c^2+ab}{a+bc} \geq 3$$
$$\iff \frac{a^2+bc}{3b+3ac} + \frac{b^2+ac}{3c+3ab} + \frac{c^2+ab}{3a+3bc} \geq 1$$
$$\iff \frac{a^2+bc}{(a+b+c)b+3ac} + \frac{b^2+ac}{(a+b+c)c+3ab} + \frac{c^2+ab}{(a+b+c)a+3bc} \geq 1$$
$$\Leftarrow \frac{a^2+bc}{(a+b+c)b+(a^2+ac+c^2)} + \frac{b^2+ac}{(a+b+c)c+(a^2+ab+b^2)} + \frac{c^2+ab}{(a+b+c)a+(b^2+bc+c^2)} = 1$$
r9m
- 17,938