I am not too sure about the following proof of the Jensen Inequality, in fact, I also doubt about the proposition.
The first thing is that $f:C \to R$ so, How does they know that $\sum_{i=1}^m \lambda_ix_i \in C$, in order to apply $f$ to this vector. Later, in the proof they apply $f$ to $\sum_{i=1}^{m-1} \lambda_ix_i/(1 - \lambda_m)$, again, how does they know that this belongs to $C$.
A full translation from spanish would likely be welcomed in an english forum, but I guess it's okay. The proof checks out as far as I reviewed it, though as you point out it could have delved into more detail.
To answer: "How does they know that $∑_m^i=1λ_i x_i∈C$, in order to apply f to this vector. ": From the list of hypothesis (or from the description of entities) $C$ is convex, hence if
$$ x_1, x_2 \in C, \lambda_1,\lambda_2 \geq0, \lambda_1+\lambda_2=1 \Rightarrow \lambda_1x_1+\lambda_2 x_2 \in C $$
Think that the line connecting these two points is contained in C. Then this can be generalized for the pondered sum of $n$ points.
To answer why $ ∑^{m−1}_{i=1}λ_ix_i/(1−λ_m) \in C$, just note that the coefficient of each $x_i$ is positive yet smaller than unity.
Assuming $1 >\lambda_m > 0$, then:
$$ 1 > (1-\lambda_m) > 0 $$
Then for each $1>\lambda_i\geq0$
Assume (for contradiction) that:
$$ \frac{\lambda_i}{1-\lambda_m} \geq 1 $$ Then: $$ \lambda_i \geq 1-\lambda_m \Rightarrow \lambda_i +\lambda_m \geq 1 $$
Which either means that all other $\lambda_k$ are zero, thus amking the situation fall into the inductive step, or a contradiction of the hypothesis that $\sum \lambda_i =1$.