Question

3．已知實(shí)數(shù)x，y，z為正數(shù)，則$\frac{xy+yz}{{{x^2}+{y^2}+{z^2}}}$的最大值為$\frac{\sqrt{2}}{2}$．

Answer 1

分析 實(shí)數(shù)x，y，z為正數(shù)，則$\frac{xy+yz}{{{x^2}+{y^2}+{z^2}}}$=$\frac{\frac{x}{y}+\frac{z}{y}}{（\frac{x}{y}）^{2}+1+（\frac{z}{y}）^{2}}$，令$\frac{x}{y}$=a＞0，$\frac{z}{y}$=b＞0，a+b=t＞0．則$\frac{xy+yz}{{{x^2}+{y^2}+{z^2}}}$=$\frac{a+b}{{a}^{2}+^{2}+1}$≤$\frac{a+b}{\frac{（a+b）^{2}}{2}+1}$=$\frac{t}{\frac{{t}^{2}}{2}+1}$=$\frac{1}{\frac{t}{2}+\frac{1}{t}}$，再利用基本不等式的性質(zhì)即可得出．

解答 解：∵實(shí)數(shù)x，y，z為正數(shù)，則$\frac{xy+yz}{{{x^2}+{y^2}+{z^2}}}$=$\frac{\frac{x}{y}+\frac{z}{y}}{（\frac{x}{y}）^{2}+1+（\frac{z}{y}）^{2}}$，
令$\frac{x}{y}$=a＞0，$\frac{z}{y}$=b＞0，a+b=t＞0．
則$\frac{xy+yz}{{{x^2}+{y^2}+{z^2}}}$=$\frac{a+b}{{a}^{2}+^{2}+1}$≤$\frac{a+b}{\frac{（a+b）^{2}}{2}+1}$=$\frac{t}{\frac{{t}^{2}}{2}+1}$=$\frac{1}{\frac{t}{2}+\frac{1}{t}}$≤$\frac{1}{2\sqrt{\frac{t}{2}×\frac{1}{t}}}$=$\frac{\sqrt{2}}{2}$，當(dāng)且僅當(dāng)t=$\sqrt{2}$，即a=b=$\frac{\sqrt{2}}{2}$，即$\frac{x}{y}$=$\frac{z}{y}$=$\frac{\sqrt{2}}{2}$時(shí)取等號(hào)．
故答案為：$\frac{\sqrt{2}}{2}$．

點(diǎn)評(píng) 本題考查了換元法、基本不等式的性質(zhì)，考查了推理能力與計(jì)算能力，屬于難題．