已知x,y,z为正实数，且满足:xy+yz+zx+xyz=4.
求证 (1/x+1/y+1/z)(x+y+z)≥(16-7xyz)/xyz

已知x,y,z为正实数，且满足:xy+yz+zx+xyz=4.
求证 (1/x+1/y+1/z)(x+y+z)≥(16-7xyz)/xyz

证明 因为xy+yz+zx+xyz=4，所以可设
x=2a/(b+c),y=2b/(c+a),z=2c/(a+b).
其中a,b,c∈R+

对证不等式置换得:
[(b+c)/a+(c+a)/b+(a+b)/c][a/(b+c)+b/(c+a)+c/(a+b)]
≥[2(b+c)*(c+a)*(a+b)-7abc]/abc.

＜＝＝＞
[∑bc(b+c)]*[∑a(a+b)(a+c)]≥∏(b+c)*[2∏(b+c)-7abc]

＜＝＝＞
∑(b+c)a^5-2∑(bc)^3-2abc∑a^3-abc∑(b+c)a^2≥0

设a=min(a,b,c),上式分解为
a^2*(ab+ac+2bc)*(a-b)*(a-c)
+[a^4-2(b+c)a^3+bca^2+a(b^3+c^3)+bc(b+c)^2](b-c)^2≥0.