1.
log(3)2 = ln2/ln3 < ln2
a a
-3 < x < 4 得 13 > -2x + 7 > -1,即 13 > y > a
x > 4 时
y = x - 4 - (x + 3) = -7 > a
综上所示
a < -7
4.
x^2 + y^2 + 1 - (xy + x + y)
= ((x - y)^2)/2 + (x^2 - 2x)/2 + (y^2 - 2y)/2 +1
= ((x - y)^2)/2 + ((x-1)^2)/2 - 1/2 + ((y-1)^2)/2 - 1/2 + 1
= ((x-y)^2 + (x-1)^2 + (y-1)^2)/2
≥ 0 (当 x = y = 1 时为0)
所以
x^2 + y^2 + 1 ≥ xy + x + y