Question

1．已知正數(shù)a，b滿足a+b=2，則$\frac{1}{a+1}+\frac{4}{b+1}$的最小值為$\frac{9}{4}$．

Answer 1

分析 正數(shù)a，b滿足a+b=2，則a+1+b+1=4．利用“乘1法”與基本不等式的性質(zhì)即可得出．

解答 解：正數(shù)a，b滿足a+b=2，則a+1+b+1=4．
則$\frac{1}{a+1}+\frac{4}{b+1}$=$\frac{1}{4}$[（a+1）+（b+1）]$（\frac{1}{a+1}+\frac{4}{b+1}）$=$\frac{1}{4}$$（5+\frac{b+1}{a+1}+\frac{4（a+1）}{b+1}）$≥$\frac{1}{4}（5+2\sqrt{\frac{b+1}{a+1}×\frac{4（a+1）}{b+1}}）$=$\frac{1}{4}×（5+4）$=$\frac{9}{4}$，
當(dāng)且僅當(dāng)a=$\frac{1}{3}$，b=$\frac{8}{3}$時(shí)原式有最小值．
故答案為：$\frac{9}{4}$．

點(diǎn)評(píng) 本題考查了“乘1法”與基本不等式的性質(zhì)，考查了推理能力與計(jì)算能力，屬于中檔題．