x,y都大于零,若x+y=5,求(x+1/x)(y+1/y)的最小值?

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

x,y都大于零,若x+y=5,求(x+1/x)(y+1/y)的最小值?
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.