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Vì xyz=1\(\Rightarrow x^2\left(y+z\right)\ge2x^2\sqrt{yz}=2x\sqrt{x}\)
Tương tự \(y^2\left(z+x\right)\ge2y\sqrt{y};z^2=\left(x+y\right)\ge2z\sqrt{z}\)
\(\Rightarrow P\ge\frac{2x\sqrt{x}}{y\sqrt{y}+2z\sqrt{z}}+\frac{2y\sqrt{y}}{z\sqrt{z}+2x\sqrt{x}}+\frac{2z\sqrt{z}}{x\sqrt{x}+2y\sqrt{y}}\)
Đặt \(x\sqrt{x}+2y\sqrt{y}=a;y\sqrt{y}+2z\sqrt{z}=b;z\sqrt{z}+2x\sqrt{x}=c\)
\(\Rightarrow x\sqrt{x}=\frac{4c+a-2b}{9};y\sqrt{y}=\frac{4a+b-2c}{9};z\sqrt{z}=\frac{4b+c-2a}{9}\)
\(\Rightarrow P\ge\frac{2}{9}\left(\frac{4c+a-2b}{b}+\frac{4a+b-2c}{a}+\frac{4b+c-2a}{b}\right)\)
\(=\frac{2}{9}\text{ }\left[4\left(\frac{c}{b}+\frac{a}{c}+\frac{b}{a}\right)+\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-6\right]\ge\frac{2}{9}\left(4.3+2-6\right)=2\)
Min P =2 khi và chỉ khi a=b=c khi va chỉ khi x=y=z=1
Thay \(1=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\) ta có
\(1+x=x+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\)
Tương tự \(1+y=\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\) và \(1+z=\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{z}+\sqrt{y}\right)\)
\(\Rightarrow\sqrt{\left(1+x\right)\left(1+y\right)\left(1+z\right)}=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{z}+\sqrt{x}\right)\)
và \(\frac{\sqrt{x}}{1+x}+\frac{\sqrt{y}}{1+y}+\frac{\sqrt{z}}{1+z}\)
\(=\frac{\sqrt{x}}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{z}\right)}+\frac{\sqrt{y}}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)}+\frac{\sqrt{z}}{\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{z}+\sqrt{y}\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)+\sqrt{y}\left(\sqrt{z}+\sqrt{x}\right)+\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{z}+\sqrt{x}\right)}\)
\(=\frac{2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{z}+\sqrt{x}\right)}\)
\(=\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{z}+\sqrt{x}\right)}\)
Do đó P = 2
\(\frac{x+\left(\sqrt{x}-\sqrt{z}\right)^2}{y+\left(\sqrt{y}-\sqrt{z}\right)^2}=\frac{\left(\sqrt{x}+\sqrt{y}-\sqrt{z}\right)^2-y+\left(\sqrt{x}-\sqrt{z}\right)^2}{\left(\sqrt{x}+\sqrt{y}-\sqrt{z}\right)^2-x+\left(\sqrt{y}-\sqrt{z}\right)^2}\)
\(=\frac{\left(\sqrt{x}+2\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{z}\right)+\left(\sqrt{x}-\sqrt{z}\right)^2}{\left(2\sqrt{x}+\sqrt{y}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{z}\right)+\left(\sqrt{y}-\sqrt{z}\right)^2}\)
\(=\frac{\left(\sqrt{x}-\sqrt{z}\right)\left(2\sqrt{x}+2\sqrt{y}-2\sqrt{z}\right)}{\left(\sqrt{y}-\sqrt{z}\right)\left(2\sqrt{x}+2\sqrt{y}-2\sqrt{z}\right)}\)
\(=\frac{\sqrt{x}-\sqrt{z}}{\sqrt{y}-\sqrt{z}}\)
