Question

[hello sun]

tìm x,y,z biết 

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

Answer 1

[Hồng Phúc]

ĐK: \(x\ge-1;y\ge3;z\ge1\)

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

\(\Leftrightarrow x+1-2\sqrt{x+1}+1+y-3-2\sqrt{y-3}+1+z-1-2\sqrt{z-1}+1=0\)

\(\Leftrightarrow\left(\sqrt{x+1}-1\right)^2+\left(\sqrt{y-3}-1\right)^2+\left(\sqrt{z-1}-1\right)^2=0\)

Ta thấy: \(\left(\sqrt{x+1}-1\right)^2+\left(\sqrt{y-3}-1\right)^2+\left(\sqrt{z-1}-1\right)^2\ge0\)

Đẳng thức xảy ra khi:

\(\left\{{}\begin{matrix}\sqrt{x+1}=1\\\sqrt{y-3}=1\\\sqrt{z-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=4\\z=2\end{matrix}\right.\)

Answer 2

[Hồng Phúc]

Cách khác:

ĐK: \(x\ge-1;y\ge3;z\ge1\)

Áp dụng BĐT \(ab\le\dfrac{a^2+b^2}{2}\).

\(\sqrt{x+1}\le\dfrac{x+1+1}{2}=\dfrac{x+2}{2}\)

\(\sqrt{y-3}\le\dfrac{y-3+1}{2}=\dfrac{y-2}{2}\)

\(\sqrt{z-1}\le\dfrac{z-1+1}{2}=\dfrac{z}{2}\)

Cộng vế theo vế các BĐT trên ta được:

\(\sqrt{x+1}+\sqrt{y-3}+\sqrt{z-1}\le\dfrac{1}{2}\left(x+y+z\right)\)

Đẳng thức xảy ra khi:

\(\left\{{}\begin{matrix}\sqrt{x+1}=1\\\sqrt{y-3}=1\\\sqrt{z-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=4\\z=2\end{matrix}\right.\)