Only consider here, so see x=2√3/3.

When x

When x & gt2√3/3, f'(x)>0 f(x) increases.

That is, f(2√3/3) is the minimum value, f(0)=3, f(3)= 18.

So f(3) is the maximum.

So f (x1) ≤ (t 2) x2-12t+3 holds.

As long as (t 2) x2- 12t+3 ≥ 18.

At this time, x2 ≥ (12t+ 15)/t 2 holds.

If x2 ∈ 0,3 is known, then 0 ≥ (12t+ 15)/t 2.

12t+ 15≤0

t≤-5/4

Just what you want.