2x+5y=20
x/10+y/4=1.
∴1/x+1/y
=1·(1/x+1/y)
=(x/10+y/4)(1/x+1/y)
=(x/10y)+(y/4x)+7/20
≥2√[(x/10y)·(y/4x)]+7/20
=2√(1/40)+7/20
=(7+2√10)/20.
故最小值为:(7+2√10)/20
2x+5y=20
x/10+y/4=1.
∴1/x+1/y
=1·(1/x+1/y)
=(x/10+y/4)(1/x+1/y)
=(x/10y)+(y/4x)+7/20
≥2√[(x/10y)·(y/4x)]+7/20
=2√(1/40)+7/20
=(7+2√10)/20.
故最小值为:(7+2√10)/20