Inequality with powers of sums

Question

Let $x_i$ be real numbers with $\sum_{i=1}^N x_i = 0$.

Show the following inequality: $$ N (\sum _{i=1}^N x_i^4) (\sum _{i=1}^N x_i^2) \ge (\sum _{i=1}^N x_i^2)^3 + N (\sum _{i=1}^N x_i^3)^2 $$

Edit: Note this particular form of the Chebyshev-Korkin Identity (page 6 in the link): $$ N \sum _{i=1}^N x_i^4 = (\sum _{i=1}^N x_i^2)^2 + \sum _{i>j} (x_i^2 -x_j^2)^2 $$ This can be used to transform the question into showing that $$ (\sum _{i=1}^N x_i^2) \sum _{i>j} (x_i^2 -x_j^2)^2 \ge N (\sum _{i=1}^N x_i^3)^2 $$

My idea would be Cauchy-Schwarz but that would require $x_i \ge 0$ which is not the case.

For arbitrary $N$ we have that equality holds for $x_1 = \cdots = x_{N-1}$ and $x_N = -(N-1)x_1$ as can be seen by plugging in: $$ \scriptstyle N \cdot(N-1 + (N-1)^4) \cdot(N-1 + (N-1)^2) = (N-1+ (N-1)^2)^3 + N\cdot (N-1 -(N-1)^3)^2 $$ and observing that this is indeed an identity.

For $N=4$ equality further holds for $x_1 = x_2 = -x_3 = -x_4$. Certainly other cases can be identified.

Note that for Chebyshev's Inequality it is always true when $\sum x_i^3=0$. — user, Jan 11 '18 at 21:56
@gimusi Yes, thanks. This can be even transformed into an equality which can be used to modify the question, see my edit in the main text. — Andreas, Jan 12 '18 at 10:56

score 0 · Accepted Answer · answered Jan 16 '18 at 14:42

In this post it is proved that, for $w_i > 0$, the following matrix (all sums $\sum_{i=1}^N $): $$ A= \begin{bmatrix} \sum_i w_i^0 & \sum_i w_i & \sum_i w_i^2 \\ \sum_i w_i & \sum_i w_i^2 & \sum_i w_i^3 \\ \sum_i w_i^2 & \sum_i w_i^3 & \sum_i w_i^4 \\ \end{bmatrix} $$is positive semidefinite, since $A$ can be written as the following quadratic form: $$ \begin{aligned} A &= \begin{bmatrix} 1 & 1 & \dots & 1 \\ w_0 & w_1 & \dots & w_N \\ w_0^{2} & w_1^{2} & \dots & w_N^{2} \end{bmatrix} \begin{bmatrix} 1 & 1 & \dots & 1 \\ w_0 & w_1 & \dots & w_N \\ w_0^{2} & w_1^{2} & \dots & w_N^{2} \end{bmatrix}^T\triangleq B^TB\end{aligned} $$ Now define the mean $\mu = \frac1N \sum_{i=1}^N w_i$ and define $w_i = \mu + x_i$. Obviously $\sum_{i=1}^N x_i = 0$. Then $A$ becomes $$ A= \begin{bmatrix} N & N \mu & N \mu^2 + \sum_i x_i^2 \\ N \mu & N \mu^2 + \sum_i x_i^2 & N \mu^3 + 3 \mu\sum_i x_i^2+ \sum_i x_i^3 \\ N \mu^2 + \sum_i x_i^2 & N \mu^3 + 3 \mu\sum_i x_i^2+ \sum_i x_i^3 & N \mu^4 + 6 \mu^2 \sum_i x_i^2+ 4 \mu\sum_i x_i^3+ \sum_i x_i^4 \\ \end{bmatrix} $$ Since $A$ is positive semidefinite, we have that $\det A \ge 0$. Evaluating the determinant gives $$ \det A = N (\sum _{i=1}^N x_i^4) (\sum _{i=1}^N x_i^2) - (\sum _{i=1}^N x_i^2)^3 - N (\sum _{i=1}^N x_i^3)^2\ge 0 $$ This directly proves the claim. $\Box$

Inequality with powers of sums

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