For $k \in \mathbb{Z}_+$ and a given $A \in \mathbb{Z}_+$, I know that $\mathbb{E}\left[\widehat{f_A}\right] \leq A - \frac{A^2}{k}$, where $\widehat{f_A}$ is a random variable dependent on $A$.
Now, however, assume that $\widehat{A}$ is a random variable. Can I say that $\mathbb{E}\left[\widehat{f_{\widehat{A}}}\right] \leq \mathbb{E}\left[\widehat{A} - \frac{\widehat{A}^2}{k}\right]$ regardless of the distribution of $\widehat{A}$?
If so, what rule or proof says that this is valid? If not, can I use this inequality to relate them at all?
Edit: I might have a proof using the definition of expectation. I am given that \begin{align} \mathbb{E}\left[\widehat{f_A}\right] & = \sum_{f_A}f_A\cdot\mathbf{P}(\widehat{f_A}=f_A) \\ & \leq A - \frac{A^2}{k} \end{align}
So I know that \begin{align} \mathbb{E}\left[\widehat{f_\widehat{A}}\right] & = \sum_{A}\left(\sum_{f_A}f_A\cdot\mathbf{P}(\widehat{f_\widehat{A}}=f_A | \widehat{A}=A)\right)\mathbf{P}(\widehat{A}=A) \\ & \leq \sum_{A}\left( A - \frac{A^2}{k} \right)\mathbf{P}(\widehat{A}=A) \\ & = \mathbb{E}\left[\widehat{A} - \frac{\widehat{A}^2}{k}\right] \end{align}
Is this proof valid?