Assume that the result is false. For each number x in the sequence, form the ordered pair (i, j), where i is the length of the longest increasing subsequence beginning with x, and j is the length of the longest decreasing subsequence ending with x. Then, since the result is false, 1im and 1jn. Thus we have mn+1 ordered pairs, of which at most mn are distinct. Hence two members of the sequence, say a and b, are associated with the same ordered pair (s, t). Without loss of generality we may assume that a precedes b in the sequence.

If a<b, then a, together with the longest increasing subsequence beginning with b, is an increasing subsequence of length (s+1), contradicting the fact that s is the length of the longest increasing subsequence beginning with a. Hence ab. But then, b, together with the longest decreasing subsequence ending with a, is a subsequence of length (t+1), contradicting that the longest decreasing subsequence ending with b is of length t. There is no way out; our assumption is false, and the result is therefore true.

(There is an interactive illustration of the above result and another one, perhaps a little more entertaining.)

References

1. M. Aigner, G. Ziegler, Proofs from THE BOOK, Springer, 2000 2000
2. A. Engel, Problem-Solving Strategies, Springer Verlag, 1998, p. 61
3. M. Gardner, The Last Recreations, Copernicus, 1997