Question

若正实数a,b满足4a+5b=6，求y=1/（a+2b）+4/（3a+3b）的最小值

Answer 1

解：
4a+5b=6
(a+2b)+(3a+3b)=6
y=1/(a+2b)+4/(3a+3b)
=(1/6)[6/(a+2b)+ 24/(3a+3b)]
=(1/6)[[(a+2b)+(3a+3b)]/(a+2b) +4[(a+2b)+(3a+3b)]/(3a+3b)]
=(1/6)[1+ (3a+3b)/(a+2b) +4(a+2b)/(3a+3b) +4]
=(1/6)[(3a+3b)/(a+2b) +4(a+2b)/(3a+3b) +5]
由均值不等式得：
(3a+3b)/(a+2b) +4(a+2b)/(3a+3b)
≥2√[[(3a+3b)/(a+2b)][4(a+2b)/(3a+3b)]]
=4
y≥(1/6)(4+5)=3/2
y的最小值为3/2