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.