Formally show that $MLE_{unconstrained} \ge MLE_{constrained}$ This makes perfect sense - If a model doesn't even assume some structure on the data, then for every value of a predictor, e.g. $x_i$, the MLE is simply the point in that position $\hat \mu_i = y_i$ (or the average of all points with the same predictor).

Yet if you impose some structure on the data, say a linear line, your MLE will be shifted:

Question is - how do I formalize this in the language of Math?
 A: In general, if $B \subseteq A$, then $$\max_{x \in A} f(x) = \max \left( \max_{x \in B} f(x), \max_{x \in A \setminus B} f(x) \right) \ge \max_{x \in B} f(x)$$
So if $B$ is the set of elements in $A$ that satisfy an additional constraint, then $B \subseteq A$, and the maximum of any function $f$ over the subset $B$ cannot be larger than the maximum of $f$ over the unconstrained set $A$.
To apply this to maximum likelihood estimation, we are really trying to compare two sets of functions: $A$ is a starting set of functions, and $B$ is a subset of the functions in $A$ that satisfy an additional constraint. Then $x \in A$ is a function, and $f(x)$ is the likelihood of the observed data using the function $x$ as a model for that data. In practice, we usually parametrize these sets of functions with a parameter $\theta \in \mathbb{R}^k$ and think about $\tilde{A}$ being the set of parameters that parametrize the functions in $A$, and $\tilde{B}$ being the set of parameters that parametrize the functions in $B$.
For example, we could let $A$ be the set of polynomial functions of degree 3 or less, and we could let $B$ be the set of linear functions. Hence $B \subseteq A$. If we wanted to work with the parameters instead, we can parametrize the set of polynomials of degree 3 or less by a vector $\theta = (\beta_0, \beta_1, \beta_2, \beta_3)$ which corresponds to the function $g(t) = \beta_0 + \beta_1 t + \beta_2 t^2 + \beta_3 t^3$. Here, $\tilde{A} = \mathbb{R}^4$, and the constrained set $\tilde{B} = \{ \theta \in \tilde{A} : \beta_2 = \beta_3 = 0 \}$, and of course $\tilde{B} \subseteq \tilde{A}$.
Addendum: Note that you cannot talk about maximum likelihood estimation without defining a likelihood function, so let's think about how the likelihood function in Case A could be defined.
The picture suggests that there is some structure being imposed on the data: Gaussians don't come from the aether, so there is some assumption behind the Gaussians drawn at each x-value, and each of the Gaussians also seems to have the same standard deviation (but where does that number come from?). The Gaussians are also displayed only along the y-axis without any along the x-axis, so it seems like the picture is saying that the x-values are fixed (non-random) and given. All of these are assumptions about how we are modeling the data and should be clearly stated.
An unconstrained model is really an "every function" model rather than a "no function" model. Assuming the four $x$-values from the picture are fixed, the set of functions $A$ is all functions $f : \{ x_1, x_2, x_3, x_4 \} \to \mathbb{R}$, i.e. those with domain consisting of the $x$-values for the four given points. You could parametrize this class of function using four parameters $\mu = \{ \mu_1, \mu_2, \mu_3, \mu_4 \}$, each of which represents the output $f(x_i) = \mu_i$ of the function. To define the likelihood function, you then need to assume something about the probability of observing the given $y$-values $\{ y_1, y_2, y_3, y_4 \}$ for a given parameter $\mu$. If you want to assume that the probability density for each of the $y_i$'s is normally distributed with a fixed standard deviation $\sigma > 0$ as the picture suggests, then $p(y_i | \mu_i) = \mathcal{N}(\mu_i, \sigma)(y_i)= \frac{1}{\sigma \sqrt{2 \pi}} \exp \left[ -\left( \frac{y_i - \mu_i}{\sigma} \right)^2 \right]$. Now if we also assume independence, we can write the likelihood function as $L(\mu) = \prod_{i = 1}^4 p(y_i | \mu_i)$, which is maximized when $\mu_i = y_i$ for $i = 1, 2, 3, 4$.
Now that the assumptions are clearly stated, note that this is still a particular instance of the general claim that $B \subseteq A \Rightarrow \max_{x \in A} f(x) \ge \max_{x \in B} f(x)$, because we can write any linear function in terms of the same parameters $\mu$ (The $\mu_i$'s simply have to satisfy the constraint that the four points $(x_i, \mu_i)$ are collinear).
