设a>b>0,求a^2+1⼀ab+1⼀a(a-b)的最小值

2025-05-14 03:33:25
推荐回答(3个)
回答1:

a2+1/ab+1/a(a-b)= ab+1/ab+a(a-b)+1/a(a-b)≥4
当且仅当 ab=1/ab,a(a-b)=1/a(a-b)取等号
即 a=√2,b=√2/2取等号.
∴ a2+1/ab+1/a(a-b)的最小值为4

回答2:

a^2+1/ab+1/a(a-b)
=[a^2-ab]+[ab+1/ab]+1/a(a-b)
=[a(a-b)+1/a(a-b)]+[ab+1/ab]
≥2+2=4

回答3:

1/[b(a-b)]≥4/a 所以 2a +1/ab+1/a(a-b)-10ac+25c =2a =a +4/a + a - 10ac+25c ≥2a 2/a +(a-5c) ≥4 最小值为4