Any $n$-qubit state can be expressed as
$$\rho=\frac{1}{2^{n}} \sum_{\mu_{1}, \ldots, \mu_{n}=0,1,2,3} T_{\mu_{1}, \ldots, \mu_{n}} \sigma_{\mu_{1}} \otimes \ldots \otimes \sigma_{\mu_{n}}$$
where $\sigma_{\mu_k}\in \{\mathbb{1,\sigma_1,\sigma_2,\sigma_3\}}$. The coefficients $T_{\mu_{1}, \ldots, \mu_{n}}$ are real numbers in $[-1,1]$ given by the correlation function values for measurements of products of Pauli operators
$$T_{\mu_{1}, \ldots, \mu_{n}}=\left\langle\sigma_{\mu_{1}} \otimes \ldots \otimes \sigma_{\mu_{n}}\right\rangle_{\rho}=\operatorname{Tr}\left(\rho \sigma_{\mu_{1}} \otimes \ldots \otimes \sigma_{\mu_{n}}\right)$$
We will call $\mathcal T$ the correlation tensor.
The paper now says, that since $\text{Tr}(\rho^2_j)\leq 1$, all 1-body correlation tensors must obey
$$||T^{(j)}||\leq \sqrt{\frac{2(d_j-1)}{d_j}} $$ with equality iff the state is pure, where the number of superscripts denotes the order of the tensor, $||\cdot ||$ is the standard Euclidean norm for vectors and $d_j$ is the dimension of the Hilbert space $\mathcal{H}_j$.
Unfortunately, the paper does not argue, how this follows from the fact that the trace is bounded by $1$.
Does anyone know how this identity follows from the trace, or where such a proof can be found?
Thanks!
EDIT: thanks to @NorbertSchuch I now understand how the above holds for qubits with $d=2$. However, I am still unclear how the general case of $d>2$ is proven.