设x为正整数，则函数y＝x2−x+1x的最小值是多少？

Question

设x为正整数，则函数y＝x2−x+(1)/(x)的最小值是多少？

Answer 1

∵y=x²-x+(1)/(x)=x（x-1）+1-(x−1)/(x)=1+(x2(x−1)−(x−1))/(x)=1+((x−1)(x2−1))/(x)=1+((x−1)2(x+1))/(x)，
∵x为正整数，
∴((x−1)2(x+1))/(x)≥0，
当x=1时，((x−1)2(x+1))/(x)=0，
∴y=1+((x−1)2(x+1))/(x)≥1．
∴函数y＝x2−x+(1)/(x)的最小值是1．