Question

cho a,b >0 thỏa mãn \(\sqrt{a}\)+\(\sqrt{b}\)=1. cmr a nhân b nhân (a+b)² <= 1 phần 64. dấu bằng xảy ra khi nào

Answer 1

Áp dụng BĐT cô -si  \(\left(ab\le\frac{\left(a+b\right)^2}{4}\right)\) ta có :

\(\frac{1}{2}\cdot2\sqrt{ab}\left(a+b\right)\le\frac{1}{2}\cdot\frac{\left(a+b+2\sqrt{ab}\right)^2}{4}=\frac{1}{2}\cdot\frac{\left(\sqrt{a}+\sqrt{b}\right)^4}{4}=\frac{1}{8}\)

<=> \(\sqrt{ab}\left(a+b\right)\le\frac{1}{8}\)

<=> \(ab\left(a+b\right)^2\le\frac{1}{64}\)

Dấu '' = '' xảy ra khi a = b = \(\frac{1}{4}\)

Answer 2

BPT <=> \(\sqrt{ab}\left(a+b\right)\le\frac{1}{8}\)

\(\frac{1}{2}\cdot2\sqrt{ab}\left(a+b\right)\le\frac{1}{2}\cdot\frac{\left(a+2\sqrt{ab}+b\right)^2}{4}=\frac{1}{2}\cdot\frac{\left(\sqrt{a}+\sqrt{b}\right)^4}{4}=\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}\)