Question

已知a>1,b>1/2，且满足2ab=a+2b+1,则2a+b的最小值是

求过程

Answer 1

等式化为：2ab-a-2b+1=2
(2b-1)(a-1)=2,
令p=a-1>0, q=2b-1>0
则有pq=2
2a+b=2(p+1)+0.5(q+1)=2p+0.5q+2.5>=2√[2p*0.5q]+2.5=√2+2.5
当2p=0.5q时取等号，此时p=1/√2,q=2√2
因此2a+b的最小值为√2+2.5

Answer 2

2ab=a+2b+1
2(a-1)b=a+1
a>1 a-1>0
b=(a+1)/[2(a-1)]
2a+b=2a+ (a+1)/[2(a-1)]
=2(a-1) +(a-1+2)/[2(a-1)] +2
=2(a-1) +1/(a-1) +5/2
由均值不等式得2(a-1)+1/(a-1)≥2√2，当且仅当a=1+ √2/2时取等号
此时b=(a+1)/[2(a-1)]=(2+√2/2)/[2(√2/2)]=(2√2+1)/2=√2 +1/2>1/2，满足tiyi9。
2a+b有最小值=2√2+ 5/2