已知证书x,y满足2x+5y=20,求1/x+1/y的最小值上述证书是正数

已知证书x,y满足2x+5y=20,求1/x+1/y的最小值上述证书是正数
已知证书x,y满足2x+5y=20,求1/x+1/y的最小值
上述证书是正数

已知证书x,y满足2x+5y=20,求1/x+1/y的最小值上述证书是正数
（2x+5y）(1/x+1/y)
=7+2x/y+5y/x
≥7+2√(2x/y*5y/x)
=7+2√10
所以,(1/x+1/y)≥(7+2√10)/（2x+5y)=（7+2√10）/20
1/x+1/y的最小值为：（7+2√10）/20

把y换成x，利用常用的不等式求解

2x+5y=20
x=4
y=12/5
1/x+1/y=1/4+5/12=(3+5)/12=8/12=2/3

1/x+1/y的最小值=2/3

The answer is (7+2*sqrt(10))/20.
Using Cauthy's inequality that
(a^2+b^2)(c^2+d^2)>=(ac+bd)^2, for all a,b,c,d>0,
(where the equality holds true if and only if a/c=b/d)
we can directly...

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The answer is (7+2*sqrt(10))/20.
Using Cauthy's inequality that
(a^2+b^2)(c^2+d^2)>=(ac+bd)^2, for all a,b,c,d>0,
(where the equality holds true if and only if a/c=b/d)
we can directly derive
(2x+5y)(1/x+1/y)
>=(sqrt(2)+sqrt(5))^2
=7+2*sqrt(10),
where the equality holds if and only if 2x^2=5y^2, which is feasible.
So the minimum value is (7+2*sqrt(10))/20.

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