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Câu 2: $P=\frac{xy.\sqrt{z-4}+yz.\sqrt{x-2}+xz.\sqrt{y-3}}{xyz}$$\Rightarrow P=\frac{\sqrt{z-4}}{z}+\frac{\sqrt{x-2}}{x}+\frac{\sqrt{y-3}}{y}$Có $\sqrt{z-4}=\frac{\sqrt{4(z-4)}}{2}\leq \frac{4+z-4}{2.2}=\frac{z}{4}$$\Rightarrow \frac{\sqrt{z-4}}{z}\leq \frac{z}{4}.\frac{1}{z}=\frac{1}{4}$Tương tự $\sqrt{x-2}=\frac{\sqrt{2(x-2)}}{\sqrt{2}}\leq \frac{x}{2\sqrt{2}}$$\Rightarrow \frac{\sqrt{x-2}}{x}\leq \frac{1}{2\sqrt{2}}$$\sqrt{y-3}=\frac{\sqrt{3(y-3)}}{\sqrt{3}}\leq \frac{y}{2\sqrt{3}}\Rightarrow \frac{\sqrt{y-3}}{y}\leq \frac{1}{2.\sqrt{3}}$Vậy $max=\frac{1}{4}+\frac{1}{2.\sqrt{2}}+\frac{1}{2.\sqrt{3}}$.Dấu = xảy ra khi $z=8;x=4;y=2$

Câu 2: $P=\frac{xy.\sqrt{z-4}+yz.\sqrt{x-2}+xz.\sqrt{y-3}}{xyz}$$\Rightarrow P=\frac{\sqrt{z-4}}{z}+\frac{\sqrt{x-2}}{x}+\frac{\sqrt{y-3}}{y}$Có $\sqrt{z-4}=\frac{\sqrt{4(z-4)}}{2}\leq \frac{4+z-4}{2.2}=\frac{z}{4}$$\Rightarrow \frac{\sqrt{z-4}}{z}\leq \frac{z}{4}.\frac{1}{z}=\frac{1}{4}$Tương tự $\sqrt{x-2}=\frac{\sqrt{2(x-2)}}{\sqrt{2}}\leq \frac{x}{2\sqrt{2}}$$\Rightarrow \frac{\sqrt{x-2}}{x}\leq \frac{1}{2\sqrt{2}}$$\sqrt{y-3}=\frac{\sqrt{3(y-3)}}{\sqrt{3}}\leq \frac{y}{2\sqrt{3}}\Rightarrow \frac{\sqrt{y-3}}{y}\leq \frac{1}{2.\sqrt{3}}$Vậy $max=\frac{1}{4}+\frac{1}{2.\sqrt{2}}+\frac{1}{2.\sqrt{3}}$.Dấu = xảy ra khi $z=8;x=4;y=6$

Câu 2: $P=\frac{xy.\sqrt{z-4}+yz.\sqrt{x-2}+xz.\sqrt{y-3}}{xyz}$$\Rightarrow P=\frac{\sqrt{z-4}}{z}+\frac{\sqrt{x-2}}{x}+\frac{\sqrt{y-3}}{y}$Có $\sqrt{z-4}=\frac{\sqrt{4(z-4)}}{2}\leq \frac{4+z-4}{2.2}=\frac{z}{4}$$\Rightarrow \frac{\sqrt{z-4}}{z}\leq \frac{z}{4}.\frac{1}{z}=\frac{1}{4}$Tương tự $\sqrt{x-2}=\frac{\sqrt{2(x-2)}}{\sqrt{2}}\leq \frac{x}{2\sqrt{2}}$$\Rightarrow \frac{\sqrt{x-2}}{x}\leq \frac{1}{2\sqrt{2}}$$\sqrt{y-3}=\frac{\sqrt{3(y-3)}}{\sqrt{3}}\leq \frac{y}{2\sqrt{3}}\Rightarrow \frac{y-3}{y}\leq \frac{1}{2.\sqrt{3}}$Vậy $max=\frac{1}{4}+\frac{1}{2.\sqrt{2}}+\frac{1}{2.\sqrt{3}}$.Dấu = xảy ra khi $z=8;x=4;y=2$

Câu 2: $P=\frac{xy.\sqrt{z-4}+yz.\sqrt{x-2}+xz.\sqrt{y-3}}{xyz}$$\Rightarrow P=\frac{\sqrt{z-4}}{z}+\frac{\sqrt{x-2}}{x}+\frac{\sqrt{y-3}}{y}$Có $\sqrt{z-4}=\frac{\sqrt{4(z-4)}}{2}\leq \frac{4+z-4}{2.2}=\frac{z}{4}$$\Rightarrow \frac{\sqrt{z-4}}{z}\leq \frac{z}{4}.\frac{1}{z}=\frac{1}{4}$Tương tự $\sqrt{x-2}=\frac{\sqrt{2(x-2)}}{\sqrt{2}}\leq \frac{x}{2\sqrt{2}}$$\Rightarrow \frac{\sqrt{x-2}}{x}\leq \frac{1}{2\sqrt{2}}$$\sqrt{y-3}=\frac{\sqrt{3(y-3)}}{\sqrt{3}}\leq \frac{y}{2\sqrt{3}}\Rightarrow \frac{\sqrt{y-3}}{y}\leq \frac{1}{2.\sqrt{3}}$Vậy$max=\frac{1}{4}+\frac{1}{2.\sqrt{2}}+\frac{1}{2.\sqrt{3}}$.Dấu = xảy ra khi$z=8;x=4;y=2$