1. 問題陳述：已知 $x,y,z$ 均為正實數，且滿足條件 $$x^{2}+y^{2}+z^{2}=1$$ 求表達式 $$\sqrt{2}xy + yz$$ 的最大值。 2. 我們希望最大化函數 $$f = \sqrt{2}xy + yz$$，在約束條件 $$x^{2}+y^{2}+z^{2} = 1$$ 下。 3. 因為 $x,y,z>0$，我們可以使用拉格朗日乘數法設置： $$ L = \sqrt{2}xy + yz - \lambda (x^{2} + y^{2} + z^{2} - 1) $$ 4. 對 $x,y,z$ 求偏導並設為零： $$ \frac{\partial L}{\partial x} = \sqrt{2}y - 2\lambda x = 0 $$ $$ \frac{\partial L}{\partial y} = \sqrt{2}x + z - 2\lambda y = 0 $$ $$ \frac{\partial L}{\partial z} = y - 2\lambda z = 0 $$ 5. 從第一式，有 $$\sqrt{2}y = 2\lambda x \Rightarrow y = \frac{2\lambda x}{\sqrt{2}} = \sqrt{2}\lambda x$$。 6. 從第三式，有 $$y = 2\lambda z \Rightarrow z = \frac{y}{2\lambda}$$。 7. 用 $y = \sqrt{2}\lambda x$ 代入第二式： $$ \sqrt{2}x + z - 2\lambda y = 0 \\ \Rightarrow \sqrt{2}x + z - 2\lambda (\sqrt{2}\lambda x) = 0 \\ \Rightarrow \sqrt{2}x + z - 2\lambda \sqrt{2} \lambda x = 0 \\ \Rightarrow \sqrt{2}x + z - 2\sqrt{2} \lambda^{2} x = 0 \\ \Rightarrow z = 2\sqrt{2} \lambda^{2} x - \sqrt{2} x = \sqrt{2} x (2\lambda^{2} - 1) $$ 8. 用 $z = \frac{y}{2\lambda}$ 及 $y = \sqrt{2}\lambda x$， 代入得： $$ z = \frac{\sqrt{2}\lambda x}{2\lambda} = \frac{\sqrt{2}}{2} x = \frac{x}{\sqrt{2}} $$ 9. 將第7與第8式求解得到： $$ \sqrt{2} x (2\lambda^{2} - 1) = \frac{x}{\sqrt{2}} \\ \Rightarrow 2\lambda^{2} - 1 = \frac{1}{2} \\ \Rightarrow 2\lambda^{2} = \frac{3}{2} \\ \Rightarrow \lambda^{2} = \frac{3}{4}\Rightarrow \lambda = \frac{\sqrt{3}}{2} $$ 10. 已知 $$ y = \sqrt{2}\lambda x = \sqrt{2} \times \frac{\sqrt{3}}{2} x = \frac{\sqrt{6}}{2} x $$ $$ z = \frac{x}{\sqrt{2}} $$ 11. 代入約束條件： $$ x^{2} + y^{2} + z^{2} = x^{2} + \left(\frac{\sqrt{6}}{2} x\right)^{2} + \left( \frac{x}{\sqrt{2}} \right)^{2} = 1 $$ $$ \Rightarrow x^{2} + \frac{6}{4} x^{2} + \frac{1}{2} x^{2} = 1 $$ $$ \Rightarrow x^{2} + 1.5 x^{2} + 0.5 x^{2} = 1 $$ $$ \Rightarrow 3 x^{2} = 1 \Rightarrow x^{2} = \frac{1}{3} \Rightarrow x = \frac{1}{\sqrt{3}} $$ 12. 得到 $$ y = \frac{\sqrt{6}}{2} \times \frac{1}{\sqrt{3}} = \frac{\sqrt{6}}{2\sqrt{3}} = \frac{\sqrt{2}}{2} $$ $$ z = \frac{1}{\sqrt{3} \sqrt{2}} = \frac{1}{\sqrt{6}} $$ 13. 最後計算最大值： $$ \sqrt{2} x y + y z = \sqrt{2} \times \frac{1}{\sqrt{3}} \times \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{\sqrt{6}} = $$ $$ \frac{\sqrt{4}}{2 \sqrt{3}} + \frac{\sqrt{2}}{2 \sqrt{6}} = \frac{2}{2\sqrt{3}} + \frac{\sqrt{2}}{2\sqrt{6}} = \frac{1}{\sqrt{3}} + \frac{\sqrt{2}}{2 \sqrt{6}} $$ $$ \Rightarrow \frac{1}{\sqrt{3}} + \frac{\sqrt{2}}{2\sqrt{6}} = \frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2} $$ 答：最大值為 $$\boxed{\frac{\sqrt{3}}{2}}$$。