You can equivalently write the (square root of the) fidelity as
$$\sqrt F(\rho,\sigma) = \|\sqrt\rho\sqrt\sigma\|_1,$$
where $\|\cdot\|_1$ is the one norm, defined as
$$\|A\|_1=\operatorname{tr}(|A|)=\operatorname{tr}[\sqrt{A^\dagger A}]
= \operatorname{tr}[\sqrt{AA^\dagger}].
$$
A general property of the trace norm $\|A\|_1$ is that it equals the max of $\left|\langle U,A\rangle\right|$ maximised over all unitaries $U$. And in particular, there is always some unitary $U$ such that $\|A\|_1=\langle U,A\rangle\equiv \operatorname{tr}(U^\dagger A)$. It's not crucial to our discussion here, but this $U$ is the unitary arising from the polar decomposition of $A$.
Suppose now $P,Q$ are generic Hermitian matrices. Then for some unitary $U$ we have
$$\|\sqrt P\sqrt Q\|_1 = \langle U, \sqrt P\sqrt Q\rangle
= \langle \sqrt P U,\sqrt Q\rangle,$$
where we are using the Hilbert-Schmidt inner product: $\langle A,B\rangle\equiv \operatorname{tr}(A^\dagger B)$.
Using Cauchy-Schwarz, we get the bound
$$\|\sqrt P\sqrt Q\|_1 \le \|\sqrt P U\|_2 \|\sqrt Q\|_2
= \sqrt{\operatorname{tr}(\sqrt P UU^\dagger \sqrt P)} \sqrt{\operatorname{tr}(\sqrt Q\sqrt Q)},$$
where $\|A\|_2\equiv \sqrt{\operatorname{tr}(A^\dagger A)}$.
In conclusion
$$F(P,Q) = \|\sqrt P\sqrt Q\|_1^2 \le \operatorname{tr}(P) \operatorname{tr}(Q),$$
which gives the unit upper bound for normalized density matrices.
Furthermore, the inequality is saturated when the CS inequality is, that is, when $\sqrt P U\propto \sqrt Q$. Being $P,Q$ Hermitian, this holds iff $P$ and $Q$ have the same eigenvalues and eigenvectors. In other words, iff $P=\lambda Q$ for some $\lambda\in\mathbb{R}$. When we are using density matrices, this amounts to $\rho=\sigma$.
You can have a look at https://cs.uwaterloo.ca/~watrous/TQI for (a lot) more details.
