Question

1/5=1/a+1/b a+b最大值等于多少

Answer 1

a,b>0

= = = = = = = = =

因为 1/5=1/a +1/b,

所以 ab =5(a+b).

令 t=a+b,a,b>0.

由基本不等式,

√ab ≤(a+b)/2,

所以 ab ≤(1/4) (t^2),

即 5t ≤(1/4) (t^2),

解得 t ≥20 或 t ≤0 (舍去).

所以 a+b≥20,

即 a+b的最小值为20.

= = = = = = = = =

换元法.

基本不等式.