This has presumably been known for 150 years. Does anybody know what
the result is called?
---David
It seems you have a corollary of Mahler’s inequality.
Assume your matrix has m rows and n columns. First, note that by
dividing each element in the matrix by n, you can replace all the
arithmetic means with ordinary sums and still get the same values for
AG and GA.
You then have Mahler’s inequality:
http://www.proofwiki.org/wiki/Mahler%27s_Inequality
where the finite sequences corresponds to your columns. The theorem
states two sequences only, but from the proof in the above link we see
that its extension to an arbitrary number of sequences is trivial.
Also, there are several inequalities attributed to Mahler, so a web
search can be pretty confusing.
It can also be seen as Holder's inequality (extended
by induction to n factors):
If 1/p_1 + 1/p_2 + ... 1/p_n = 1 then
\int f_1 f_2 ... f_n d\mu \le
(\int f_1^{p_1} d\mu)^{1/p_1}...(\int f_n^{p_n} d\mu)^{1/p_n},
Apply this with all p_j equal n, f_k(j) = x_{kj}^{1/n} and
\mu the probability measure assigning weight 1/n to each
j in {1,2,...,n}.
Dan
To reply by email, change LookInSig to luecking
The proofs above show AG(A) <= GA(A). It seems interesting to test out
if AG(A) <= GA(A*), where A* is the transpose of the matrix A, is also
true, ie take the geometric and arithmetic means of the COLUMNS
first.
Of course this cannot be true for general matrices (take a 1xn
matrix), and it is also false for square matrices, but just barely. A
numerical experiment shows that it is true more than 95% of the time
for square matrices with at most 10 rows, and in fact appears to be
asymptotically true as the number of rows increases. Is this a known
result?