Question

已知a＞0,b＞0,a+b=2,则y=1/a+4/b的最小值是多少?

Answer 1

y=1/a+4/b

=1/2*2*(1/a+4/b)

=1/2*(a+b)*(1/a+4/b)

=1/2*[（√a）^2+(√b)^2]*{[√(1/a)]^2+[√(4/b)]^2}

≥1/2*{[(√a) *√(1/a)]+ [(√b)* √(4/b)]}^2 (根据柯西不等式)

=1/2*(1+2)^2

=9/2

所以y的最小值是9/2