Question

已知x>0,y>0,且x^2+y^2/2=1,求x根号(1+y^2)的最大值

Answer 1

设x=cosα，y=√2sinα，0＜α＜π/2
x√(1+y^2)=cosα√[1+2(sinα)^2]
=√[(cosα)^2+2(sinαcosα)^2]
=√{(cosα)^2+(1/2)[sin(2α)]^2}
=√{(1/2)[1+cos(2α)]+(1/2)[sin(2α)]^2}
=√(1/2)√{1+cos(2α)+[sin(2α)]^2}
=√(1/2)√{2+cos(2α)-[cos(2α)]^2}
=√(1/2)√{-[cos(2α)-1/2]^2+9/4}
0＜α＜π/2
0＜2α＜π
-1＜cos(2α)＜1
-3/2＜cos(2α)-1/2＜1/2
0≤[cos(2α)-1/2]^2＜1/4或0≤[cos(2α)-1/2]^2＜9/4
0≤[cos(2α)-1/2]^2＜9/4
-9/4＜-[cos(2α)-1/2]^2≤0
0＜9/4-[cos(2α)-1/2]^2≤9/4
0＜√{9/4-[cos(2α)-1/2]^2}≤3/2
0＜x√(1+y^2)≤(3/2)√(1/2)
最大值(3/2)√(1/2)

Answer 2

x√(1+y^2)<=√2[x√(1/2+y^2/2)]<=√2/2*(x^2+1/2+y^2/2)=√2/2*(1+1/2)=3√2/4
中间用到2ab<=a^2+b^2