Question

设a>b>c>0,则2a^2+1/ab+1/a(a-b)-10ac+25c^2的最小值

b(a-b)<=(a/2)^2

取得等号的条件是，a=5c，a=2b，且a=根号2

这是怎样的到的
请详解
谢谢

Answer 1

由b(a-b)<=(a/2)^2=a^2/4可得
2a^2+1/ab+1/a(a-b)-10ac+25c^2
=a^2+1/ab+1/a(a-b)+a^2-10ac+25c^2
=a^2+1/b(a-b)+a^2-10ac+25c^2
>=a^2+4/a^2+(a-5c)^2
>=4
等号成立当且仅当b=a-b,a=5c,a^=2.

Answer 2

2a^2+1/ab+1/a(a-b)-10ac+25c^2
=a^2-10ac+25c^2+a^2+1/ab+1/a(a-b)
=(a-5)^2+(ab+1/ab)-ab+a^2+1/a(a-b)
>=(a-5)^2+2*根号(ab*1/ab)-a(a-b)+1/a(a-b)=0+2+2=4
当且仅当ab=1/ab和a(a-b)=1/a(a-b)时