问题
填空题
如果存在实数x，y，z，使得x＞y＞z，且
|
答案
x＞y＞z，且
+1 x-y
≤1 y-z
成立，两边同乘以x-z得a z-x
(x-z)(
+1 x-y
)≤-a，而(x-z)(1 y-z
+1 x-y
)=[(x-y)+(y-z)](1 y-z
+1 x-y
)=2+1 y-z
+y-z x-y
≥2+2x-y y-z
=4，当且仅当
•y-z x-y x-y y-z
=y-z x-y
，即x-y=y-z时取得等号．x-y y-z
所以4≤-a，即a≤-4，a的最大值是-4．
故答案为：-4．