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