Inequality with sum of rows of symmetric matrices Let S be a symetric matrix, with coefficients positive or zero, and T its square $$T=S^2$$
Let $S_i$ and $T_i$ be the sum of the $i$-th row (or column) of $S$ and $T$ respectively.
I noticed that the following inequality seemed to always be true. How can it be proved?
$$
  \sum_{k=1}^{n} \frac{T_i}{S_i} \geq \sum_{k=1}^{n} {S_i}
$$
Example:
$S = \begin{bmatrix}1& 2\\ 2& 3\end{bmatrix}$, so that $T = \begin{bmatrix}5&8\\ 8& 13\end{bmatrix}$. We can verify that:
$$
   \frac{5 + 8}{1 + 2} + \frac{8 + 13}{2 + 3} = 8.5333.. \geq (1 + 2) + (2 + 3) = 8
$$

As $S$ is symmetric, we have $\color{red}{\sum_j s_{kj}=S_k}$ and
$$
\sum_i \frac{T_i}{S_i}
=\sum_i \frac{\sum_j\sum_k s_{ik}s_{kj}}{S_i}
=\sum_i \frac{\sum_k (s_{ik} \color{red}{\sum_j s_{kj}})}{S_i}
=\sum_{i,k} \frac{s_{ik} S_k}{S_i}.\tag{1}
$$
Since 
$\dfrac{s_{ik} S_k}{S_i}+\dfrac{s_{ki} S_i}{S_k}=\left(\dfrac{S_k}{S_i}+\dfrac{S_i}{S_k}\right)s_{ik}\ge 2s_{ik}=s_{ik}+s_{ki}$, the result follows.
