根据题意得x+y≠0
1∕X+1∕Y=1∕X+Y
=>(x+y)/xy=1/(x+y)
=>(x+y)²=xy
因为(x+y)²≥|2xy|
所以(x+y)²>xy
即(x+y)²=xy不成立,所以原式不成立