x根号(1+y^2)＝根号{2[x^2(1+y^2)/2]}
由基本不等式a+b>=2√ab知,ab<=[(a+b)/2]^2
x^2(1+y^2)/2<=[(x^2+1/2+y^2/2)/2]^2=[(1/2+1)/2]^2=9/16
所以：根号{2[x^2(1+y^2)/2]}<=根号[2×9/16]=(3/4)根号2