Eigenvalues of density operator under quantum channel Consider a 2x2 density operator $\rho$ with eigenvalues $\lambda_{1},\lambda_{2}$ (where $\lambda_{1}\geq\lambda_{2}$) evolving into a 4x4 density operator $\rho'$ (with eigenvalues $\lambda'_{1},\lambda'_{2},\lambda'_{3},\lambda'_{4}$ in non-ascending order) under a quantum channel, 
$$\rho \to \rho'=\sum_{i} M_{i}\rho M^{\dagger}_{i}
\quad\text{where}\quad
\sum_{i}M^{\dagger}_{i}M_{i}=I.$$
I am interested in the upper bounds on the largest eigenvalue $\lambda'_{1}$ of the final density operator $\rho'$ in terms of the eigenvalues $\lambda_{1},\lambda_{2}$ of the initial density operator $\rho$. Intuitively, I think $\lambda'_{1}\leq \lambda_{1}$ must hold true, but I want to show this rigorously. 
Since I am considering a physical evolution $\rho \to \rho'$, I believe that the von-Neumann entropy can only increase i.e $S(\rho')\geq S(\rho)$. Please correct me if am wrong on this. 
However as I mentioned, I am interested in obtaining an upper bound on $\lambda'_{1}$ which seems hard to obtain from the entropy inequality. I would be grateful if someone can point me to some useful references on the spectral evolution of density operators under quantum channels or give any hints/directions.
 A: I doubt you'll be able to get useful bounds without restricting the class of channels.
Consider the following quantum channel (for arbitrary dimension $d,n$):
$$ T:\mathbb{C}^{d\times d}\to \mathbb{C}^{n\times n}: \rho \mapsto \operatorname{tr}(\rho) \sigma $$
where $\sigma$ is any quantum state. Using the Choi-Jamiolkowski isomorphism, you can easily see that this is indeed a completely positive map and by construction, it is trace-preserving.
Clearly, the spectrum of $\sigma$ is completely independent of the spectrum of $\rho$.

Okay, so there is no way to have these inequalities for arbitrary channels. You ask now:

Since I am considering a physical evolution $\rho\to \rho^{\prime}$, I believe that the von-Neumann entropy can only increase i.e $S(\rho)\to S(\rho^{\prime})$.

This is wrong. For instance, this would mean that you could never prepare pure ancillary states for a quantum computation (and you can do this at least to very good precision). It would also mean that most completely positive, trace-preserving maps would not be physical. But of course, there are large classes of processes in nature, where (local) entropy can only increase. These channels are in some sense very "dissipative".
That means that although they are not as physically all-encompassing as you ask, let's consider quantum channels with increasing entropy.
The case of equal input and output dimensions
I'm talking about this case, because I actually know the answer and maybe my comments help you to find what you need.
It turns out that a channel if entropy increasing if and only if it is unital (and therefore doubly-stochastic). That unitality is necessary for an entropy increasing channel is clear here, because the maximally mixed state has maximal entropy. The other direction follows from the fact that relative entropy only decreases under doubly-stochastic maps.
In this case, here is an answer to your question with the upshot:
For a unital quantum channel $T$, the decreasingly ordered spectrum of $T(\rho)$ is majorized by the ordered spectrum of $\rho$ or put differently
$$ \sum_{i=1}^k \lambda_i^{\downarrow}(T(\rho))\leq \sum_{i=1}^k \lambda_i^{\downarrow}(\rho) \qquad \forall k=1,\ldots,n $$
This entails your inequality for this case, but it seems you are more interested in general channels.
The case of larger output dimensions
Unlike the first case, I have never worked with this case or seen it being used in your context, so this will only be a weak attempt at an answer.
A classification is not as straightforward, because I believe I can always embed a quantum channel $T:\mathbb{C}^{2\times 2}\to \mathbb{C}^{2\times 2}$ into a larger one, where I don't do anything on the other two dimensions and this would still be a (now nonunital) entropy increasing channel. Therefore, while unital maps should still be entropy decreasing, now the converse is definitely not true.
However, if we suppose the channel is not merely entropy-increasing, but unital as above, I think I can provide you with a proof of an even stronger statement - please check for mistakes, I haven't done these manipulations in a while:
Proposition: Let $T:\mathbb{C}^{2\times 2}\to \mathbb{C}^{4\times 4}$ be a doubly-stochastic quantum channel. Then $\lambda_1(T(\rho))+\lambda_2(T(\rho))\leq \lambda_1(\rho)$ [the eigenvalues are ordered decreasingly].
The proof uses the theorem for channels with equal in- and output dimensions. Consider any $\rho$ and denote the eigenvalues of $T(\rho)$ by $\lambda_1,\ldots,\lambda_4$. In the basis, where $T$ is diagonal, we can decompose $\mathbb{C}^{4\times 4}=\mathbb{C}^{2\times 2}\otimes \mathbb{C}^{2\times 2}$ writing (in qubit notation):
$$ \sum_{i=1}^4 \lambda_i |i\rangle\langle i|=\lambda_1|0\rangle\langle 0|+\lambda_2|01\rangle\langle 01|+\lambda_3|10\rangle\langle 10|+\lambda_4|11\rangle\langle 11|$$
Till now, nothing has happened (just changing bases). Now, instead of considering the map $T$, we consider the map $\operatorname{tr}_2\circ T$, the composition of the map and the partial trace on the second subsystem. Since the partial trace is doubly-stochastic and CP, this composition is also doubly-stochastic and CP and our majorization criterion holds.
We have: 
$$\operatorname{tr}_2(T(\rho))=(\lambda_1+\lambda_2)|0\rangle\langle 0|+(\lambda_3+\lambda_4)|1\rangle\langle 1|$$
and then, via the majorization criterion: $\lambda_1+\lambda_2\leq \lambda_1(\rho)$ if $\lambda_1(\rho)$ is the largest eigenvalue of the initial state $\rho$.
With similar reasoning, it should be possible to get a number of interesting inequalities for similar questions at least for unital maps. I have no idea how to proceed for non-unital maps. Likewise, I have no real idea whether these criteria will be sufficient (but you didn't ask that anyways).
