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It is a well known fact that the fidelity is preserved by unitary evolution, i.e. $$ F(\rho ,\sigma) = F(U\rho U^\dagger, U\sigma U^\dagger), $$ for any unitary operator $U$.

However in most quantum algorithms, we also use measurements to for instance uncompute an ancillary register. Imagine for instance that we want to implement a given unitary process $U$, however an ancillary qubit is necessary to perform that operation: $$ V \left(|x\rangle |0\rangle\right) = \left(U |x\rangle\right) |+\rangle, $$ therefore we can measure the last register and discard its value. If we write $\mathcal{A}$ the described process, it is true that: $$ F(\rho ,\sigma) \le F(\mathcal{A}(\rho), \mathcal{A}(\sigma))? $$

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Yes. Your last statement is true. Fidelity is monotonic under the action of CPTP maps.

However, you can also use multiplicativity of fidelity, which is - $$ F\bigg(\big( \rho_A \otimes \rho_B \big) , \big( \sigma_A \otimes\sigma_B \big) \bigg) = F\big(\rho_A,\sigma_A\big) \cdot F\big(\rho_B,\sigma_B\big)\,. $$ In your case, it becomes -

$$\begin{align} F\bigg(\big( |x\rangle \otimes |0\rangle \big) , \big( U|x\rangle \otimes |+\rangle \big) \bigg) &= F\big(|x\rangle,U|x\rangle\big) \cdot F\big(|0\rangle,|+\rangle\big)\,,\\ &= \big|\langle x|U |x\rangle\big|^2 \cdot \big|\langle 0|+\rangle\big|^2\,,\\ &= \frac{1}{2}\cdot \big|\langle x|U |x\rangle\big|^2\,. \end{align} $$

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