分类：教育/科学-学习帮助
问题：11号全国数学竞赛了,拜求一道不等式证明.设x,y,z为正实数,? 
11号全国竞赛了,拜求一道不等式证明.设x,y,z为正实数,且满足x+y+z=1.求证1/(2-3yz)+1/(2-3zx)+1/(2-3xy)≤9/5
回答：设x,y,z为正实数,且满足x+y+z=1.求证 
1/(2-3yz)+1/(2-3zx)+1/(2-3xy)≤9/5

简证 所证不等式齐次化,等价于
Σ(x+y+z)^2/[2(x+y+z)^2-3yz]≤9/5    (1)

(1)<===>
5(Σx)^2*Σ[2(x+y+z)^2-3zx]*[2(x+y+z)^2-3xy]≤
9[2(x+y+z)^2-3yz]*[2(x+y+z)^2-3zx]*[2(x+y+z)^2-3xy] (2)

(2)展开为
12Σx^6+24Σ(y+z)x^5-12Σ(y^2+z^2)x^4-48Σ(yz)^3
+xyz[45Σx^3+15Σ(y+z)x^2-189xyz]>=0  

上式约去3得:
4Σx^6+8Σ(y+z)x^5-4Σ(y^2+z^2)x^4-16Σ(yz)^3
+xyz[15Σx^3+5Σ(y+z)x^2-63xyz]>=0           (3)

设x=min(x,y,z),(3)式分解为
x[4x^3+12(y+z)x^2+8x(y^2+z^2)+35xyz+28yz(y+z)](x-y)(x-z)
+[4(y+z)^4+31xyz(y+z)+8(y+z)x^3-(4y^2+4z^2-17yz)-8(y+z)x^3](y-z)^2>=0.
上式显然成立.