Điều kiện: \(x\ge2;y\ge-2024;z\ge2025\)
Ta có \(\sqrt{x-2}=\sqrt{1.\left(x-2\right)}\le\dfrac{1+x-2}{2}=\dfrac{x-1}{2}\) (bđt Cô-si)
\(\sqrt{y+2024}=\sqrt{1.\left(y+2024\right)}\le\dfrac{1+y+2024}{2}=\dfrac{y+2025}{2}\)
\(\sqrt{z-2025}=\sqrt{1.\left(z-2025\right)}\le\dfrac{1+z-2025}{2}=\dfrac{z-2024}{2}\)
Cộng theo vế 3 bđt trên, ta có:
\(VP=\sqrt{x-2}+\sqrt{y+2024}+\sqrt{z-2025}\)
\(\le\dfrac{x-1}{2}+\dfrac{y+2025}{2}+\dfrac{z-2024}{2}\)
\(=\dfrac{1}{2}\left(x+y+z\right)=VP\)
Như vậy dấu "=" phải xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-2=1\\y+2024=1\\z-2025=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-2023\\z=2026\end{matrix}\right.\) (nhận)
Vậy pt đã cho có nghiệm \(\left(x,y,z\right)=\left(3,-2023,2026\right)\)