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}}}
And we 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, and you end up with the formula
>>4 has as your best possible answer.