已知实数xy满足x-y≤0,x+y-5≥0,y-3≤0,若不等式(y2-axy+2x2)/x2≥y/x恒成立
问题描述：
已知实数xy满足x-y≤0,x+y-5≥0,y-3≤0,若不等式(y2-axy+2x2)/x2≥y/x恒成立
则实数a的取值范围是?(y2为y的平方)
答
x-y≤0,x+y-5≥0,y-3≤0
x≤y≤3,x+y≥5
2≤x≤y≤3
(y2-axy+2x2)/x2≥y/x
y^2-axy+2x^2≥xy
y^2-(a+1)xy+2x^2≥0
[-(a+1)]^2-4*2≤0
-1-2√2≤a≤-1+2√2