设x>y>z,n∈N,且1/(x-y)+1/(y-z)≥n/(x-z)恒成立,则n的最大值为?
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设x>y>z,n∈N,且1/(x-y)+1/(y-z)≥n/(x-z)恒成立,则n的最大值为?
设x>y>z,n∈N,且1/(x-y)+1/(y-z)≥n/(x-z)恒成立,则n的最大值为?
设x>y>z,n∈N,且1/(x-y)+1/(y-z)≥n/(x-z)恒成立,则n的最大值为?
设x>y>z,n∈N,
则1/(x-y)+1/(y-z)≥(1+1)²/[(x-y)+(y-z)] (柯西不等式)
=4/(x-z)
要使1/(x-y)+1/(y-z)≥n/(x-z)恒成立
只需4/(x-z)≥n/(x-z)
所以n的最大值为4