Question

求y(x)=(1+x平方)/4(x-1),x＞1的最小值,要求必须用均值不等式解答,正确答案是当x=3时y最小=2

Answer 1

你是不是表述有误啊,我按照你的答案把题目改了一下.

y(x)=(1+x)^2/4(x-1)

=[2+(x-1)]^2/4(x-1)

=[4+4(x-1)+(x-1)^2]/4(x-1)

=1/(x-1)+1+(x-1)/4

≥2√(1/4)+1

=2

仅当1/(x-1)=(x-1)/4时,即x=3时,等号成立,取到最小值2.