Diễn Đàn MathScope - Xem bài viết đơn - $\frac{a^2b}{4-bc}+\frac{b^2c}{4-ca}+\frac{c^2a}{4-ab} \leq 1$

dangthanhbn · 14-03-2015, 05:20 PM

Cho $\left\{\begin{matrix} a,b,c\geq 0 & \\ a+b+c=3 & \end{matrix}\right.$. Chứng minh rằng:
$\frac{a^2b}{4-bc}+\frac{b^2c}{4-ca}+\frac{c^2a}{4-ab} \leq 1$

14-03-2015, 05:20 PM	#1
dangthanhbn +Thành Viên+ Tham gia ngày: Mar 2015 Bài gởi: 1 Thanks: 0 Thanked 0 Times in 0 Posts	$\frac{a^2b}{4-bc}+\frac{b^2c}{4-ca}+\frac{c^2a}{4-ab} \leq 1$ Cho $\left\{\begin{matrix} a,b,c\geq 0 & \\ a+b+c=3 & \end{matrix}\right.$. Chứng minh rằng: $\frac{a^2b}{4-bc}+\frac{b^2c}{4-ca}+\frac{c^2a}{4-ab} \leq 1$