Question

不等式,求最值不等式:a＞b＞c＞0,求2a²＋1/ab＋1/a（a－b）-10ac＋25c²的最小

Answer 1

原式=a^2-10ac+(5c)^2+a^2-ab+ab+1/(ab)+1/(a^2-ab)

=(a-5c)^2+[ab+1/(ab)]+[a^2-ab+1/(a^2-ab)] ≥0+2+2

等号成立条件是b=1/a =a-b a=根号2 b=1/根号2