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\(\left(1\right)\Rightarrow\hept{\begin{cases}\sqrt{x}-\sqrt{y}=\frac{1}{\sqrt{z}}-\frac{1}{\sqrt{y}}=\frac{\sqrt{y}-\sqrt{z}}{\sqrt{xy}}\\\sqrt{y}-\sqrt{z}=\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{z}}=\frac{\sqrt{z}-\sqrt{x}}{\sqrt{xz}}\\\sqrt{z}-\sqrt{x}=\frac{1}{\sqrt{y}}-\frac{1}{\sqrt{x}}=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\end{cases}\left(2\right)}\)
\(\left(2\right)\Rightarrow\left(\sqrt{x}-\sqrt{y}\right).\left(\sqrt{y}-\sqrt{z}\right).\left(\sqrt{z}-\sqrt{x}\right)=\frac{\left(\sqrt{y}-\sqrt{z}\right).\left(\sqrt{z}-\sqrt{x}\right).\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{zyzxxy}}\left(3\right)\)\(Từ\left(3\right)\)Ta sẽ chứng minh được rằng \(\orbr{\begin{cases}x=y=z\\x.y.z=1\end{cases}}\)
\(P=\left(\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)+\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)}{1-xy}\right):\left(\frac{x+y+2xy+1-xy}{1-xy}\right)\)
\(=\left(\frac{2\sqrt{x}+2y\sqrt{x}}{1-xy}\right):\left(\frac{\left(x+1\right)\left(y+1\right)}{1-xy}\right)\)
\(=\frac{2\sqrt{x}\left(y+1\right)}{\left(1-xy\right)}.\frac{\left(1-xy\right)}{\left(x+1\right)\left(y+1\right)}=\frac{2\sqrt{x}}{x+1}\)
\(x=\frac{2}{2+\sqrt{3}}=\frac{2\left(2-\sqrt{3}\right)}{4-3}=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}-1\)
\(\Rightarrow P=\frac{2\left(\sqrt{3}-1\right)}{5-2\sqrt{3}}=\frac{2+6\sqrt{3}}{13}\)
Ta có \(1-P=1-\frac{2\sqrt{x}}{x+1}=\frac{x-2\sqrt{x}+1}{x+1}=\frac{\left(\sqrt{x}-1\right)^2}{x+1}\ge0\) \(\forall x\ge0\)
\(\Rightarrow1-P\ge0\Rightarrow P\le1\)