f(s？ -2s)≥-f(2t-t？ )=f(t？ -2t)

Incremental function: s? -2s≥t？ -2t

(s-t)(s+t-2)≥0

Let 3t+s = k.

t=(k-s)/3

[s-(k-s)/3][s+(k-s)/3-2]≥0

(4s-k)(2s+k-6)≥0

g(x)=8s？ -2(k- 12)s-k(k-6)≥0

g( 1)=8-2k+24-k(k-6)≥0

k？ -4k-32≤0

-4≤k≤8 ( 1)

g(4)=-k？ +2k+224≥0

- 14≤k≤ 16 (2)

What is the difference between (1) (2):-4 ≤ K ≤ 8 and your answer?

And I haven't considered δ and symmetry axis, which means the actual range is smaller.

If you follow the upstairs method, although it is consistent with the correct answer, the problem-solving process is very flawed.