设a＞b＞c＞0，则2a2+1ab+1a(a−b)-10ac+25c2的最小值是．

Question

设a＞b＞c＞0，则2a²+(1)/(ab)+(1)/(a(a−b))-10ac+25c²的最小值是______．

Answer 1

∵a＞b＞c＞0，∴2a2+1ab+1a(a−b)-10ac+25c2=a2+1b(a−b)+(a−5c)2≥a2+1(b+a−b2)2+(a−5c)2=a2+4a2+（a-5c）2≥2a2•4a2+0=4．当且仅当a=2b=5c=2时取等号．因此2a2+1ab+1a(a−b)-10ac+25c2的最小值是4．故答案为...