Let $\sum_{j=0}^{n-1}x_{i,j}=1$ $\forall i \in \{1,2,\cdots,k\}$.
How to show that $\sum_{i=1}^{k}\sum_{j=0}^{n-1}x_{i,j}^{2} \geq \frac{k}{n}$ and that we have equality if and only if $x_{i,j}=\frac{1}{n} \forall i,j$.
I am trying to understand a theorem that uses it. It seems that it is standard, but I was not able to find the original source.
Thank you in advance.
If $\sum_j x_{i,j}=1$ for every $i$ then check that $$ \left(\sum_{j=0}^{n-1} x_{i,j}^2\right) -\frac1n=\sum_{j=0}^{n-1}\left(x_{i,j}-\frac1n\right)^2.\tag1 $$ Sum (1) over all $i=1,\ldots k$ to obtain $$ \sum_{i=1}^k\left(\sum_{j=0}^{n-1} x_{i,j}^2\right) -\frac kn=\sum_{i=1}^k\sum_{j=0}^{n-1}\left(x_{i,j}-\frac1n\right)^2.\tag2$$ The RHS of (2) is the sum of non-negative terms, so it is non-negative. This proves the inequality $$\sum_{i=1}^k\left(\sum_{j=0}^{n-1} x_{i,j}^2\right) \ge\frac kn.\tag3 $$ And equality holds if and only if the RHS of (2) equals zero, which holds iff every summand on the RHS of (2) is zero.