Question

4．已知正數(shù)a，b滿足2a²+b²=3，求a$\sqrt{^{2}+1}$的最大值．

Answer 1

分析 由題意和基本不等式可得a$\sqrt{^{2}+1}$=$\frac{\sqrt{2}}{2}$•$\sqrt{2}$a•$\sqrt{^{2}+1}$≤$\frac{\sqrt{2}}{2}$•$\frac{（\sqrt{2}a）^{2}+（\sqrt{^{2}+1}）^{2}}{2}$，代值計(jì)算注意等號(hào)成立的條件可得．

解答 解：∵正數(shù)a，b滿足2a²+b²=3，
∴a$\sqrt{^{2}+1}$=$\frac{\sqrt{2}}{2}$•$\sqrt{2}$a•$\sqrt{^{2}+1}$
≤$\frac{\sqrt{2}}{2}$•$\frac{（\sqrt{2}a）^{2}+（\sqrt{^{2}+1}）^{2}}{2}$
=$\frac{\sqrt{2}}{4}$（2a²+b²+1）=$\sqrt{2}$
當(dāng)且僅當(dāng)$\sqrt{2}$a=$\sqrt{^{2}+1}$即a=b=1時(shí)取等號(hào)．
∴a$\sqrt{^{2}+1}$的最大值為$\sqrt{2}$．

點(diǎn)評(píng) 本題考查基本不等式求最值，湊出可以基本不等式的形式是解決問(wèn)題的關(guān)鍵，屬基礎(chǔ)題．