Minimizing...

A friend asked me this math question (presumably popped out of QFT or something)

Consider a set of N elements: {a_i}
It is known that sum(a_i) = 0
and that sum(a_i²) = 1
What is the minimum of sum(a_i⁴)?

The AM-RMS/power mean inequality tells us that sqrt(sum(a_i⁴)/N) >= sum(a_i²)/N ->
sum(a_i⁴) >= 1/N, with equality iff the a_is are the same.

If N is even, then the a_is just alternate as +-1/sqrt(N)
If N is odd, then that trick doesn't work, since then the sum won't = 0.  Therein lies the problem.  It seems likely that the answer has (N+1)/2 of the a_is equaling one number, and the remainder equaling something else.  However, I haven't been able to prove that this is the optimal solution yet.
I tried brute forcing this for N=3, but after an hr of algebra... the answer is always 1/2, regardless of how I pick the a_is.

Since I haet subscripts, let the numbers be a,b,c,\ldots.  Define the Lagrangian as

\Lambda(a,b,c,...,\lambda_1, \lambda_2) = (a^4 + b^4 + c^4 + \cdots) - \lambda_1 (a+b+c+\cdots) - \lambda_2 (a^2 + b^2 + c^2 + \cdots - 1)

Setting the partials with respect to a,b,c,\ldots to zero gives

4a^3 - \lambda_1 - 2\lambda_2 a = 0
4b^3 - \lambda_1 - 2\lambda_2 b = 0
4c^3 - \lambda_1 - 2\lambda_2 c = 0
...

Therefore each number a,b,c,\ldots is a root of the cubic equation f(x) = 4x^3 - \lambda_1 - 2\lambda_2 x = 0.  This shows there are at most three distinct numbers among the a,b,c,\ldots in an optimal solution.

Assume that there are exactly three distinct numbers, a,b,c, and let their multiplicities be s,t,(n-s-t) respectively.  By looking at coefficients of the cubic f(x) = 4x^3 - \lambda_1 - 2\lambda_2 x = 0, we can see that

a + b + c = 0
ab+bc+ca = -\lambda_2 / 2
a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca) = \lambda_2

-\lambda_2/2 = ab+bc+ca = ab+c(a+b) = ab+(-a-b)(a+b) = -a^2-ab-b^2
\lambda_2 = 2(a^2+ab+b^2) = 2(b^2+bc+c^2) = 2(c^2+ca+a^2)

and by looking at the original problem, we have

sa + tb + (n-s-t)c = 0
sa^2 + tb^2 + (n-s-t)c^2 = 1

We're going to use this to get expressions for a^2, b^2, c^2 similar to what >>4 did. 

Now subtract

(n-s-t)(a+b+c)=0

from the first of these last two equations and

(n-s-t)(a^2+b^2+c^2)=(n-s-t)\lambda_2 = (n-s-t)(2a^2+2ab+2b^2)

from the second, to get

(2s+t-n)a + (2t+s-n)b = 0
(2s+t-n)a + (2t+s-n)b = 1-(n-s-t)(2a^2+2ab+2b^2)

Solving for b in the first of these and subbing into the second, then solving the second for a^2 (all in Maple), gives

{a}^{2}={\frac { \left( -2t+n-s \right) ^{2}}{ \left( s+t \right) {n
}^{2}+ \left( -{s}^{2}-{t}^{2}-10ts \right) n+9st \left( s+t
 \right) }}

Similarly, we find

{b}^{2}={\frac { \left( -2s+n-t \right) ^{2}}{ \left( s+t \right) {n
}^{2}+ \left( -{s}^{2}-{t}^{2}-10ts \right) n+9\,st \left( s+t
 \right) }}

{c}^{2}={\frac { \left( -t+s \right) ^{2}}{ \left( 9t-n \right) {s}^
{2}+ \left( n-t \right)  \left( n-9t \right) s+nt \left( n-t
 \right) }}

And finally, from this we get (all in Maple)

sa^4 + tb^4 + (n-s-t)c^4 = {\frac {{n}^{2}-3ns-3tn+3{s}^{2}+3ts+3{t}^{2}}{9s{t}^{2}+9t{s}^{2}+s{n}^{2}-n{s}^{2}-n{t}^{2}+t{n}^{2}-10stn}}

Denote the RHS of this as M(s,t).  With Maple again, we find that

M_s (s,t) = -{\frac { \left( n-3t \right) ^{3} \left( -2s+n-t \right) }{
 \left( 9s{t}^{2}+9t{s}^{2}+s{n}^{2}-n{s}^{2}-n{t}^{2}+t{n}^{2}-10stn \right) ^{2}}}

M_t (s,t) = -{\frac { \left( n-3s \right) ^{3} \left( -2t+n-s \right) }{ \left( 9s{t}^{2}+9t{s}^{2}+s{n}^{2}-n{s}^{2}-n{t}^{2}+t{n}^{2}-10stn \right) ^{2}}}

You can see both partials are zero iff t = s = n/3, which gives m(s,t) = 3/2n, but I think this is a saddle point or something.  I forget how to check that.

For the boundaries, we're looking at the region bounded by s=0, t=0, s+t=n, and if you go through and check it, the minimums occur at s=0, t=n/2, s=n/2, t=0, and s=n/2, t=n/2.  So the minimum happens when there are actually only two distinct values, and each occurs the same amount of times.  In the case that n is odd, the next best thing is to let s=(n-1)/2, t=(n+1)/2, u=0, and you end up with the formula >>4 has as your best possible answer.