设x大于等于1,则函数y＝（x+2)(x+3)/x+1的最小值

设x大于等于1,则函数y＝（x+2)(x+3)/x+1的最小值
设x大于等于1,则函数y＝（x+2)(x+3)/x+1的最小值

设x大于等于1,则函数y＝（x+2)(x+3)/x+1的最小值
设x+2=t,t∈（3,+∞）,y=t(t+1)/(t-1)=(t^2+t)/(t-1)=[(t-1)^2+3t-1]/(t-1)=t-1+(3t-1/t-1)=t-1+(3t-3+2/t-1)=t-1+2/(t-1)+3≥2+2/2+3=6,(因为t-1≥2),所以函数的最小值为6.