Question

cho a,b > 0 , \(\sqrt{a}+\sqrt{b}=1\)

Hãy chứng minh ab(a+b)² ≤ \(\frac{1}{64}\)

GIÚP MÌNH VỚI Ạ!

2 ngày nữa thi ròiii

Answer 1

@Trần Khánh Hoài1h25' rồi mà vẫn thao thức hả ban:))

thi lớp 10 à

Answer 2

\(\left(\sqrt{a}+\sqrt{b}\right)=1\Leftrightarrow\sqrt{ab}=\frac{1-\left(a+b\right)}{2}\)

\(ab\left(a+b\right)^2=[\sqrt{ab}\left(a+b\right)]^2=[\frac{1-\left(a+b\right)}{2}.\left(a+b\right)]^2=[\frac{\left(a+b\right)-\left(a+b\right)^2}{2}]^2\)

ta có \(\left(a+b\right)-\left(a+b\right)^2=-\left[\left(a+b\right)^2-\left(a+b\right)+\frac{1}{4}-\frac{1}{4}\right]=-\left(a+b-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

do đó \(\frac{\left(a+b\right)-\left(a+b\right)^2}{2}\le\frac{1}{8}\Rightarrow\left[\frac{\left(a+b\right)-\left(a+b\right)^2}{2}\right]^2\le\frac{1}{64}\)