Question

已知a＞0,b＞0,a+b=2,求y=1/a+4/b的最小值

Answer 1

y=1/a +4/b

=[(1/2)(a+b)]/a +2(a+b)/b

=(1/2)(1+ b/a) +2(a/b +1)

=(1/2)(b/a) +2(a/b) +5/2

由均值不等式得,当(1/2)(b/a)=2(a/b)时,即a/b=1/2时,(1/2)(b/a)+2(a/b)有最小值2

此时,y有最小值ymin=2+ 5/2=9/2