初等證明。感覺反而比分析方法複雜。

It suffices to prove the statement for the case x, y, z>=0.

The statement is equivalent to

x^2y^2z^2 + 2{(xy-1)^2+(yz-1)^2+(zx-1)^2}+ x^2+y^2+z^2-2xy-2yz-2zx+2 >= 0, but


x^2y^2z^2 + 2{(xy-1)^2+(yz-1)^2+(zx-1)^2}+ x^2+y^2+z^2-2xy-2yz-2zx+2 >= x^2y^2z^2 + 2{(xy-1)^2+(yz-1)^2+(zx-1)^2} -xy-yz-zx+2 (because x^2+y^2+z^2 >= xy + yz + zx).

Write a=xy-1, b=yz-1, c=zx-1, the right hand side of the above equation is simplified to:

abc + ab + bc + ca + 2(a^2+b^2+c^2), which is >= 0 when a, b, c >= 0. If one of a, b, or c is negative, we know it is no less than -1. Suppose -1 <= a <= 0.

abc + ab + bc + ca + 2(a^2 + b^2 + c^2) >= abc + (a^2+b^2+c^2) >= |bc| + (a^2 + b^2 + c^2) >= 0

所有跟帖： 

• 老班主真是功力非凡。不情之請，能不能看一下前麵的85，86？ -wxcfan123- ♂ (0 bytes) () 02/12/2015 postreply 09:31:40

• Not easy. I tried No. 85 -亂彈- ♂ (0 bytes) () 02/12/2015 postreply 16:23:57

• 好。還有另一個別致的思路，也不錯 -魁北克人- ♂ (58 bytes) () 02/12/2015 postreply 13:43:14

• 這個很好。用判別式方法也是很自然的，但中間放縮等都不宗易 -亂彈- ♂ (0 bytes) () 02/12/2015 postreply 16:26:10