Question

设a,b属于R,a²+2b²=6,则b/a-3的最大值

Answer 1

设a+b＝t，则a＝t-b...............[1]
代入条件得：(t-b)^2+2b^2=6,
3b^2-2tb+(t^2-6)=0...............[2]
∵b是实数，∴判别式δ≥0,
即4t^2-12(t^2-6)≥0,
化简得：t^2≤9，
∴-3≤t≤3.
当t＝-3时，由[2]得b=-1,代入[1]得a=-2.
所以a+b的最小值是-3(当a=-2,b=-1时取到）.

Answer 2

令b/(a-3)=t
b=(a-3)t
所以
a²+2[(a-3)t]²=6
a²+2(a²t²-6at²+9t²)=6
(1+
2t
²)a²-12t²a+18t²-6=0
判别式
=144t^4-4(1+2t²)(18t²-6)≥0
144t^4-4(18t²-6+36t^4-12t²)≥0
18t²-6-12t²≤0
6t²≤6
t²≤1
-1≤t≤1
所以
b/a-3的最大值为1