因为x>0,y>0,x^2+y^2/2=1 ==>x^2+(y^2+1)/2=3/2
x√（1+y^2)=√2*x*√[(y^2+1)/2]≤√2*[x^2+(y^2+1)/2]/2=√2*(3/2)/2
=3√2/4
当x^2+y^2/2=1
x=√[(y^2+1)/2] 即x=√3/2 b=√2/2时 x√（1+y^2)有最小值3√2/4