I want to prove that $$\operatorname{Var}\left(\sum\limits_{i=1}^m{X_i}\right) \leq m\sum\limits_{i=1}^m{\operatorname{Var}(X_i)} \,. \>$$
A too complicated proof is to write $$ a_{ij}=\sqrt {Cov(X_i,X_j)} \,, $$
$$\operatorname{Var}\left(\sum\limits_{i=1}^m{X_i}\right) = \sum a_{ij}^2 \leq \sum a_{ii}a_{jj} \leq \sum\sum a_{ii}^2$$
By Cauchy-Scwarz and then the permutation inequality.
I'm sure it can be shorter, but how?
The inequality is true in general under the assumption $\textrm{Var}(X_i)<\infty$ for all $i$ (otherwise both sides are infinite). Applying Cauchy-Schwarz to the random quantities $X_i-\mathbb E X_i$ yields that $$ \mathbb P\left[\left(\sum_{i=1}^m X_i-\mathbb E X_i\right)^2\leq m\sum_{i=1}^m (X_i-\mathbb E X_i)^2\right]=1. $$ Consequently, $$ \mathbb E\left(\sum_{i=1}^m X_i-\mathbb EX_i\right)^2\leq m\ \mathbb E\sum_{i=1}^m (X_i-\mathbb EX_i)^2. $$ Using the definition of variance, we see this is exactly the desired inequality.
A simpler method is to use the inequality $a_ia_j \leq \frac 12 (a_i^2+a_j^2)$ after Cauchy-Schwarz.
