设x+2y=a,x-2y=b,则x=(a+b)/2,y=(a-b)/4,则题目可以变换为:a>0,b>0,ab=4,求|(a+b)/2|-|(a-b)/4|的最小值,因为a>0,b>0,所以(a+b)/2肯定大于0,接下来分两种可能1:a大于等于b。
|(a+b)/2|-|(a-b)/4|=(a+3b)/4,最小值为根号3,当且仅当a=3b时成立
2:a小于b,则|(a+b)/2|-|(a-b)/4|=(3a+b)/4,最小值为根号3,当且仅当b=3a等号成立
追问a大于等于b。
|(a+b)/2|-|(a-b)/4|=(a+3b)/4,最小值为根号3,当且仅当a=3b时成立
为什么?= =|||
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