Question

Tìm x,y,z

\(\sqrt{x-a}+\sqrt{y-b}+\sqrt{z-c}=\dfrac{1}{2}\left(x+y+z\right)\) (Với x+y+z=3)

Answer 1

\left\{\begin{matrix}
x+y=0\\
x-1=0\end{matrix}\right.

Answer 2

Lời giải:
Áp dụng BĐT AM-GM:

\(\sqrt{x-a}=\sqrt{1(x-a)}\leq \frac{1+x-a}{2}\)

\(\sqrt{y-b}=\sqrt{1(y-b)}\leq \frac{1+y-b}{2}\)

\(\sqrt{z-c}=\sqrt{1(z-c)}\leq \frac{1+z-c}{2}\)

Cộng theo vế:

\(\sqrt{x-a}+\sqrt{y-b}+\sqrt{z-c}\leq \frac{x+y+z+3-(a+b+c)}{2}=\frac{x+y+z+3-3}{2}=\frac{1}{2}(x+y+z)\)

Dấu "=" xảy ra khi \(\left\{\begin{matrix} x-a=1\\ y-b=1\\ z-c=1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=a+1\\ y=b+1\\ z=c+1\end{matrix}\right.\)

Vậy.......