若实数x 、y满足x>y>0且1/x-y+8/x+2y=1则x+y的最小值为
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解令x+y=m(x-y)+n(x+2y)
则m+n=1，2n-m=1,
解得n=1/3，n=2/3
故x+y=m(x-y)+n(x+2y)
=2/3(x-y)+1/3(x+2y)
=[2/3(x-y)+1/3(x+2y)]×1
=[2/3(x-y)+1/3(x+2y)]×[1/(x-y)+8/(x+2y)]
=2/3+8/3+16/3(x-y)/(x+2y)+1/3(x+2y)1/(x-y)
≥10/3+2根16/3×1/3
=10/3+8/3
=6
故函数的最小值为6.
则m+n=1，2n-m=1,
解得n=1/3，n=2/3
故x+y=m(x-y)+n(x+2y)
=2/3(x-y)+1/3(x+2y)
=[2/3(x-y)+1/3(x+2y)]×1
=[2/3(x-y)+1/3(x+2y)]×[1/(x-y)+8/(x+2y)]
=2/3+8/3+16/3(x-y)/(x+2y)+1/3(x+2y)1/(x-y)
≥10/3+2根16/3×1/3
=10/3+8/3
=6
故函数的最小值为6.
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