x,y都大于零,若x+y=5,求(x+1/x)(y+1/y)的最小值?
1个回答

解法一:

x、y>0,且

5=x+y≥2√(xy)→√(xy)≤5/2.

依Cauchy不等式,得

(x+1/x)(y+1/y)

≥[√(xy)+1/√(xy)]^2

=(5/2+2/5)^2

=841/100.

故所求最小值为:841/100.

解法二:

构造下凸函数f(t)=ln(t+1/t),则

f(x)+f(y)≥2f[(x+y)/2]

→ln(x+1/x)+ln(y+1/y)≥2ln[(x+y)/2+2/(x+y)]

→ln[(x+1/x)(y+1/y)]≥ln(5/2+2/5)^2

→(x+1/x)(y+1/y)≥841/100.

故所求最小值为:841/100.