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文[1]对求函数y=x2+(2-x/x-1)2(x≥0,且x≠1)的最小值,采用了如下一种利用基本不等式的求解方法:“y=x2+(2-x/x-1)2=(x-1)2+(2-x/x-1)2+2x-1≥2((x-1)2(2-x/x-1)2)1/2+2x-1=2(2-x)+2x-1=3.(当且仅当x-1=2-x/x-1,即x=51/2+1/2时等号成立).”
In [1], we use the following method to solve the minimum inequality y = x2 + (2-x / x-1) 2 (x≥0 and x ≠ 1) (2-x / x-1) 2 = (x-1) 2+ (2-x / x-1) 2 + 2x- 1≥2 ((x- 1) 2 (If and only if x-1 = 2-x / x-1, ie x = 51/2 + 1/2) 2 when the equal sign is established). "