Question

若正数a、b满足1/a+4/b=2，则a+b的最小值为多少？

Answer 1

解析：
∵1/a+4/b=2,∴a=b/(2b-4),
则a+b=b/(2b-4)+b=(2b^2-3b)/(2b-4)
=[2(b-2)^2+5(b-2)+2]/2(b-2)
=(b-2)+1/(b-2)+5/2,
∵1/a+4/b=2,a,b为正，∴b＞2,
即b-2＞0,,
∴a+b≥2√(b-2)*[1/(b-2)]+5/2=9/2,
当且仅当b-2=1/(b-2),即b=3,取=号，
则a+b的最小值=9/2

Answer 2

解：∵a,b＞0
1/a+4/b=2
∴1/2a+2/b=1
∴a+b=(a+b)（1/2a+2/b）
=1/2+2a/b+b/2a+2
=5/2+2a/b+b/2a
≥5/2+2=9/2
∴a+b的最小值为9/2

Answer 3

1/a+4/b=2
4a+b = 2ab
a+b = 2ab -3a --(1)

let S = a+b
b= S-a ---(2)

from （1） and (2)
S = 2a(S-a) -3a
S = (2a^2 + 3a) / (2a-1)
S'= (4a+3)/(2a-1) - 2(2a^2 + 3a)/(2a-1)^2
S' =0
=> (4a+3)(2a-1) - 2(2a^2 + 3a) =0 (a is not equal to 1/2)
8a^2+2a-3 - 4a^2-6a=0
4a^2-4a-3=0
(2a-3)(2a+1)=0
a= 3/2 or -1/2 (rejected)

S''(3/2) > 0 (min)

min (a+b) at a =3/2
min (a+b) = (9/2 + 9/2)/(3-1)
= 9/2 #