x^2+(y^2)/2=1,
x^2+[(1/√2)y]^2=1,
設(shè)x=cosA,y=√2sinA,
因x＞0,y＞0,不妨設(shè)0＜A＜π/2,
x√(1+y^2)=cosA√[1+2(sinA)^2]
=√{(cosA)^2[1+2(sinA)^2]}
=√{[1-(sinA)^2][1+2(sinA)^2]}
=√[1+(sinA)^2-2(sinA)^4]
=√{1+(1/√2)[√2(sinA)^2]-[√2(sinA)^2]^2}
=√{-[√2(sinA)^2-1/(2√2)]^2+[1/(2√2)]^2+1}
=√{-[√2(sinA)^2-√2/4]^2+9/8}
=√{-2[(sinA)^2-1/4]^2+9/8}
=√{-2[(sinA+1/2)(sinA-1/2)]^2+9/8}
≤√(9/8)=(3/4)√2
當(dāng)sinA=1/2,即x=y=√2/2時(shí),原式取最大值(3/4)√2.