Question

如果存在实数x，y，z，使得x＞y＞z，且

1

x-y

+

1

y-z

≤

a

z-x

成立，则实数a的最大值是______．

Answer 1

x＞y＞z，且

1

x-y

+

1

y-z

≤

a

z-x

成立，两边同乘以x-z得
(x-z)(

1

x-y

+

1

y-z

)≤-a，而(x-z)(

1

x-y

+

1

y-z

)=[(x-y)+(y-z)](

1

x-y

+

1

y-z

)=2+

y-z

x-y

+

x-y

y-z

≥2+2

y-z x-y • x-y y-z

=4，当且仅当

y-z

x-y

=

x-y

y-z

，即x-y=y-z时取得等号．
所以4≤-a，即a≤-4，a的最大值是-4．
故答案为：-4．