Question

Chứng minh rằng:

\(\sqrt{c.\left(a-c\right)}+\sqrt{c.\left(b-c\right)}\le\sqrt{ab}\) với a>0, b>0, c>0

Answer 1

Áp dụng bđt Bunhiacopxki :

\(\sqrt{c}\cdot\sqrt{a-c}+\sqrt{c}\cdot\sqrt{b-c}\le\sqrt{\left[\left(\sqrt{c}\right)^2+\left(\sqrt{a-c}\right)^2\right]\left[\left(\sqrt{c}\right)^2+\left(\sqrt{b-c}\right)^2\right]}\)

\(=\sqrt{\left(c+a-c\right)\left(c+b-c\right)}=\sqrt{ab}\) ( đpcm )

Dấu "=" xảy ra \(\Leftrightarrow\frac{c}{a-c}=\frac{c}{b-c}\Leftrightarrow a-c=b-c\Leftrightarrow a=b\)