[x(√2-y)]^2=x^2(√2-y)^2=(8-4y^2)(√2-y)^2=4(√2+y)(√2-y)^3=(4/3)*3(√2+y)(√2-y)^3≤(4/3)*{[3√2+3y+√2-y+√2-y+√2-y]/4}^4=3^3即max=3√3
已知x^2/8+y^2/2=1 求x(√2-y)的最大值.
[x(√2-y)]^2=x^2(√2-y)^2=(8-4y^2)(√2-y)^2=4(√2+y)(√2-y)^3=(4/3)*3(√2+y)(√2-y)^3≤(4/3)*{[3√2+3y+√2-y+√2-y+√2-y]/4}^4=3^3即max=3√3