Some power-sum inequalities ===Warren D. Smith====July 2015==== Given a vector x of N positive reals, define the E-power-sum of x to be F_E(x) = SUM(0<j<=N) (x_j)^E where E is any real number. As a trivial example, F_0(x)=N. THEOREM: Let x be an N-vector of positive reals. If B<0<A or A<0<B then F_A(x) * F_B(x) >= N * F_{A+B}(x). If 0<=min(A,B) then F_A(x) * F_B(x) <= N * F_{A+B}(x). All these inequalities become equalities if all the x_j are equal, for example if all x_j=1. WHO SHOULD GET THE CREDIT? Normally I would doubt that this result could be new. However, I then found Bruce Reznick: Some inequalities for products of power sums, Pacific J. Maths 104,2 (1983) 443-463 http://projecteuclid.org/euclid.pjm/1102723674 which appears to prove a weaker result via a much lengthier argument. So maybe this actually is a new result (?). Perhaps by return mail I will receive news this is Foobuznik's inequality proved in 1785. EXAMPLE: if A=B=1 then the following inequality holds for all vectors x of nonnegative reals: F_1(x) * F_1(x) <= N * F_2(x) since this is a form of the well known "root mean square inequality." And that will imply the more general inequality F_A(x) * F_A(x) <= N * F_{2A}(x) is valid for any real number A. PROOF: Our first QUESTION is: for which A and B does an inequality between F_A(x) * F_B(x) and N * F_{A+B}(x) hold? 1. We can wlog demand A+B=1 because if A+B=C is not 1, then just redefine x_j to be (x_j)^C, and then (A, B, C) are replaced by (A/C, B/C, 1) whereupon the validity (or non-validity) of the inequality is unaffected. 2. We also may wlog then demand that the sum F_1(x)=1, since if it equalled any other value (say S) we could rescale x by replacing x_j by x_j / S, and again the validity (or non-validity) of the inequality would be unaffected. 3. Observe that the function F_A(x) is a concave-U function of the vector x if either A>1 or A<0, but concave the other way if 0<A<1. That is since the power function mapping z to z^A is concave-U if A>1 or if A<0, for all z>0; and since any sum of concave-U functions is concave-U. 4. The global min of such a concave-U function necessarily occurs when all the x_j are equal. And this includes the global min within the hyperplane F_1(x)=1, since restricting a concave function to a hyperplane, still is concave. 5. And as we remarked before, the inequality we are trying to prove, obviously is an equality under this "x_j all equal" circumstance. 6. Therefore: if A>1 or A<0, then among vectors x with F_1(x)=1, the unique vector minimizing F_A(x) has all entries equal. Similarly for F_B(x). Therefore: F_A(x) * F_B(x) >= N * F_1(x) if A+B=1 with A>1 or B>1, and hence F_A(x) * F_B(x) >= N * F_{A+B}(x) if B<0<A or A<0<B. And similarly from the concavity in the other direction (and maxxing rather than minning) F_A(x) * F_B(x) <= N * F_1(x) if A+B=1 with 0<=A and 0<=B, and hence F_A(x) * F_B(x) <= N * F_{A+B}(x) if 0<=min(A,B). Q.E.D. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)