Bài toán :
Cho các số thưc không âm $a,b,c$ thoả mãn $ab+bc+ca=1$. Chứng minh rằng :
$$\dfrac{(1+ab)^2}{a^2+b^2+4ab}+\dfrac{(1+bc)^2}{b^2+c^2+4bc}+\dfrac{(1+ca)^2}{c^2+a^2+4ca} \ge \dfrac{8}{3}$$
$$\dfrac{(1+ab)^2}{a^2+b^2+4ab}+\dfrac{(1+bc)^2}{b^2+c^2+4bc}+\dfrac{(1+ca)^2}{c^2+a^2+4ca} \ge \dfrac{8}{3}$$
Started By Tham Lang, 22-08-2012 - 21:07
#1
Posted 22-08-2012 - 21:07
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