设实数X,Y,A,B,满足X^2+Y^2=3,A^2+B^2=1,则AX+BY的最大值是?我的思路:X^2+Y^2=3-----------------------①A^2+B^2=1-----------------------②①+②得:X^2+Y^2+A^2+B^2=4X^2+A^2≥2XAY^2+B^2≥2YB所以,2XA+2YB≥4所以,AX+BY≥4但是,=
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