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ĐK: \(x;y;z\ge0;x+z\ge y.\)
\(\Leftrightarrow\sqrt{x-y+z}+\sqrt{y}=\sqrt{x}+\sqrt{z}\)Bình phương 2 vế >=0
\(\Leftrightarrow2\sqrt{y}\sqrt{x+z-y}=2\sqrt{x}\sqrt{y}\Leftrightarrow\sqrt{y\left(x+z-y\right)}=\sqrt{xz}\)Bình phương 2 vế >=0
\(\Leftrightarrow y\left(x+z-y\right)=xz\Leftrightarrow y\left(z-y\right)+x\left(y-z\right)=0\Leftrightarrow\left(z-y\right)\left(y-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=y\\z=y\end{cases}}\)
Vậy đẳng thức đề bài cho được thỏa mãn với mọi x = y hoặc z = y. (và \(x;y;z\ge0;x+z\ge y.\)).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{x-y+z}=\sqrt{x}-\sqrt{y}+\sqrt{z}\)
Điều kiện tự làm nhé
\(\Leftrightarrow x-y+z=x+y+z+2\left(\sqrt{xz}-\sqrt{xy}-\sqrt{yz}\right)\)
\(\Leftrightarrow y+\sqrt{xz}-\sqrt{xy}-\sqrt{yz}\)
\(\Leftrightarrow\left(\sqrt{z}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=y\\z=y\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(1.\sqrt{1+x^2}+1.\sqrt{2x}\le\sqrt{\left(1+1\right)\left(1+x^2+2x\right)}=\sqrt{2}\left(x+1\right)\)
Tương tự:
\(\sqrt{1+y^2}+\sqrt{2y}\le\sqrt{2}\left(y+1\right)\) ; \(\sqrt{1+z^2}+\sqrt{2z}\le\sqrt{2}\left(z+1\right)\)
Cộng vế:
\(P\le\sqrt{2}\left(x+y+z+3\right)+\left(2-\sqrt{2}\right)\left(x+y+z\right)\le\sqrt{2}\left(3+3\right)+\left(2-\sqrt{2}\right).3=6+3\sqrt{2}\)
\(P_{max}=6+3\sqrt{2}\) khi \(x=y=z=1\)
Bình phương 2 vế, ta đc:\(x-y+z=x+y+x-2\sqrt{xy}-2\sqrt{yz}+2\sqrt{zx}\Rightarrow y-\sqrt{xy}-\sqrt{yz}+\sqrt{zx}=0\Rightarrow\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}-\sqrt{z}\right)=0\)Tự lm nốt nha.
\(\sqrt{x-y+z}=\sqrt{x}-\sqrt{y}+\sqrt{z}\Leftrightarrow\left(\sqrt{x-y+z}\right)^2=\left(\sqrt{x}-\sqrt{y}+\sqrt{z}\right)^2\Leftrightarrow x-y+z=x+y+z-2\sqrt{xy}-2\sqrt{yz}+2\sqrt{xz}\Leftrightarrow2y-2\sqrt{xy}-2\sqrt{yz}+2\sqrt{xz}=0\Leftrightarrow y-\sqrt{xy}-\sqrt{yz}+\sqrt{xz}=0\Leftrightarrow\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)-\sqrt{z}\left(\sqrt{y}-\sqrt{x}\right)=0\Leftrightarrow\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}-\sqrt{z}\right)=0\Leftrightarrow\left[{}\begin{matrix}\sqrt{y}-\sqrt{x}=0\\\sqrt{y}-\sqrt{x}=0\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}\sqrt{y}=\sqrt{x}\\\sqrt{y}=\sqrt{z}\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}y=x\\y=z\end{matrix}\right.\)