m(2)= 3(6 & lt； 7 & lt9)，m(3)= 3(5 & lt； 7 & lt9)，m(4)= 2(7 & lt； 9)，m(5)= 1 (9)，m(6)= 2(3 & lt； 8)，m(7)= 2(2 & lt； 8)，m(8)= 1 (8)，m(9)= 1 ( 1)

So m( 1)=4, and the maximum value of m(i) is 4.

(2) 4 7 2 9 1 6 8 3 5。

(3) On the contrary, if there is no increase or decrease subsequence of at least t+ 1, let's push the contradiction angrily.

Let m(i) represent the length of the longest monotonically increasing subsequence starting from a(i). According to the disproof condition that m (I) < =t, at the same time, since the length of the increasing sequence of at least a(i) is 1, M (I) >: = 1. That is to say, there are t cases in which the value that m(i) can take * * * ranges from 1 to t.

The number t 2+ 1 has t values from m( 1) to m (t 2+ 1). According to the pigeon hole principle, there are at least [(t 2+ 1)/t]+ 1.

Let m (b (1)) = m (b (2)) = ... = m (b (t+ 1)) is the same number of t+1,where b (1).

Let's prove that a (b (1)) >: A(b(2)), because if a (b (1))

Similarly, it can be inferred that a (b (1)) > a(b(2))& gt； a(b(3))& gt； ... & gtA(b(t+ 1)), because b (1) ...

This is the end of the proof. Let's call it a day.