Inequality with powers of sums 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.  
 A: 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. $\Box$
