考點(diǎn):基本不等式
專(zhuān)題:不等式的解法及應(yīng)用
分析:實(shí)數(shù)x>0,y>0,且x
2+y
2=1,可設(shè)x=cosθ,y=sinθ,
θ∈(0,).于是(x+
)(y+
)=x+y+
+=
sinθ+cosθ+,令sinθ+cosθ=t=
sin(θ+)∈
(1,],可得(x+
)(y+
)=t+
=t+
-=f(t),利用導(dǎo)數(shù)研究其單調(diào)性即可得出.
解答:
解:∵實(shí)數(shù)x>0,y>0,且x
2+y
2=1,
∴可設(shè)x=cosθ,y=sinθ,
θ∈(0,).
則(x+
)(y+
)=x+y+
+=sinθ+cosθ+
+=
sinθ+cosθ+,
令sinθ+cosθ=t=
sin(θ+)∈
(1,],
∴t
2=1+2sinθcosθ,解得sinθcosθ=
,
∴(x+
)(y+
)=t+
=t+
-=f(t),
∴f′(t)=1-
--
=
<0,
∴函數(shù)f(t)在
t∈(1,]上單調(diào)遞減,
∴當(dāng)t=
時(shí),函數(shù)f(t)取得最小值,
f()=
+1.
故答案為:
+1.
點(diǎn)評(píng):本題考查了“三角函數(shù)代換”方法、利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性極值、同角三角函數(shù)基本關(guān)系式,考查了推理能力與計(jì)算能力,屬于難題.