I'll reproduce here a standard argument used to prove the fundamental bound for pretty good measurements (PGMs), the the most part taken from Watrous' book, with some minor changes in notation, presentation, and an attempt at deriving the structure of PGMs without knowing it a priori (that however still requires to know part of the structure of the solution, so this approach only goes a little bit in that direction). I don't find this to be a particularly satisfactory answer to the question about "the idea behind" pretty good measurements, but being this bound arguably the main reason PGMs are interesting, I figured it's still good to have this derivation here, for reference.
Proof idea -
The main trick of the derivation, the way I see it, is applying a "double CS inequality" after rewriting the inner product that corresponds to the outcome probabilities.
It's a bit like saying: we want to estimate the inner product $\langle \mathbf v,\mathbf w\rangle\equiv \sum_i v_i w_i$. We can rewrite it as
$\langle\mathbf v,\mathbf w\rangle = \langle P \mathbf v,P^{-1}\mathbf w\rangle$ for some invertible Hermitian $P$. Using the CS inequality then gives a different bound for each choice of $P$:
$$\langle \mathbf v,\mathbf w\rangle \le \|P\mathbf v\| \|P^{-1}\mathbf w\|.$$
Problem setting - Say we want to discriminate between an ensemble of states $a\mapsto (p_a,\rho_a)$, meaning the possible states are $\{\rho_a\}_a$, and the $a$-th state is known to occur with probability $p_a$. Define the ensemble operators as $\eta_a\equiv p_a \rho_a$, and $\eta\equiv \sum_a \eta_a$, so that $\operatorname{tr}(\eta)=\sum_a \operatorname{tr}(\eta_a)=1$.
Let $a\mapsto \mu_a$ be an arbitrary POVM with number of outcomes equal to the number of elements in the ensemble (we can assume this wlog). A standard way to use the POVM to discriminate between the states, is to guess the input state to being $\rho_a$ if the $a$-th outcome is found. This happens with probability $p_a=\langle\mu_a,\eta_a\rangle\equiv \operatorname{tr}(\mu_a\eta_a)=p_a \operatorname{tr}(\mu_a\rho_a)$.
It follows that the overall probability of correctly guessing the input state using this POVM and strategy is
$$p_{\rm guess}(\mu) = \sum_a \langle \mu_a,\eta_a\rangle
= \sum_a p_a \langle \mu_a,\rho_a\rangle.$$
Ideally, we'd like to find the optimal discrimination strategy, that is, maximise $p_{\rm guess}(\mu)$ over the set of possible POVMs. This amounts to a semidefinite program. However, the point of pretty good measurements (PGMs) is to find a recipe to find a measurement strategy that works well enough, but doesn't require performing such optimisation to find it.
Derivation - To find such strategy, we can try using the following trick: let $A$ be some invertible Hermitian positive definite matrix; then
$$\langle \mu_a, \eta_a\rangle = \langle A\mu_a A,A^{-1}\eta_a A^{-1}\rangle.$$
This is immediate to verify writing the inner product with the trace. We also don't actually need $A$ to be invertible. It is sufficient to assume $\operatorname{im}(\eta_a)\subseteq \operatorname{im}(A)$, and use the pseudoinverse instead of the actual inverse. This ensures $A^{+}\eta_a A^{+}$ is well-defined even though $A$ is not invertible, where $A^+$ indicates the standard Moore-Pensores pseudoinverse.
From this:
Using CS, we get for every $a$, $$\langle A\mu_a A,A^{+}\eta_a A^{+}\rangle
\le \| A\mu_a A\|_2 \|A^+ \eta_a A^+\|_2
= \langle A^2 \mu_a A^2,\mu_a\rangle \langle A^{+ 2}\eta_a A^{+2},\eta_a\rangle$$
From the above we have
$$\sum_a \langle \mu_a,\eta_a\rangle \le
\sum_a \| A\mu_a A\|_2 \|A^+ \eta_a A^+\|_2.$$
We can think of this as again the inner product between the two vectors $(\| A\mu_a A\|_2 )_a$ and $( \|A^+ \eta_a A^+\|_2 )_a$, and thus using CS again,
$$\sum_a \langle \mu_a,\eta_a\rangle \le
\left(\sum_a \| A\mu_a A\|_2^2\right)^{1/2}
\left(\sum_b \| A^+\eta_b A^+\|_2^2\right)^{1/2}.$$
Now looking into the two terms we got, note that for the first one
$$\sum_a \| A\mu_a A\|_2^2 =
\sum_a \langle A\mu_a A,A\mu_a A\rangle
= \sum_a \operatorname{tr}( A^2 \mu_a A^2 \mu_a)
\\
\le \sum_a \operatorname{tr}(A^4 \mu_a) = \operatorname{tr}(A^4),$$
where we used $\operatorname{tr}(PQ)\le \operatorname{tr}(P)$ if $P,Q\ge0$ and $Q\le I$, and the standard normalisation condition $\sum_a \mu_a = I$.
On the other hand, the other term reads
$$ \sum_b \| A^+\eta_b A^+\|_2^2 =
\sum_b \langle A^{+2} \eta_b A^{+2},\eta_b\rangle. $$
Remember that what we are looking for is an upper bound for $p_{\rm guess}(\mu)$ in terms of the guessing probability $p_{\rm guess}(\mu^{\rm PGM})$ of some POVM $\mu^{\rm PGM}$ that only depends on the ensemble $a\mapsto \eta_a$ (and has a "sufficiently simple" expression).
Looking at the bound we got for $p_{\rm guess}(\mu)$, we've got it, if we can find $A$ such that $\operatorname{tr}(A^4)=1$, and such that $(A^{+2}\eta_b A^{+2})_b$ is a POVM. Given that the only information we're allowed to use is the ensemble $a\mapsto \eta_a$, and that $\operatorname{tr}(\eta)=1$, it's pretty natural to try defining $A=\eta^{1/4}\equiv\left(\sum_a \eta_a\right)^{1/4}$, and thus
$$\mu^{\rm PGM}_b \equiv \sqrt{\eta^+} \eta_b \sqrt{\eta^+},$$
and it's simple enough to check that this is indeed a valid POVM.
We thus showed that using the POVM $\mu^{\rm PGM}_b \equiv \sqrt{\eta^+} \eta_b \sqrt{\eta^+}$ the discrimination probability satisfies the bound
$$p_{\rm guess}(\mu^{\rm PGM})^2 \ge p_{\rm guess}(\mu), \,\,\forall \mu,$$
and thus also $p_{\rm guess}(\mu^{\rm PGM})^2 \ge \max_\mu p_{\rm guess}(\mu)$.
