Let $f:\mathbb R^d\to\mathbb R\cup\{+\infty\}$ be a proper convex lower semicontinuous function. Suppose that $f$ is bounded by below, and for simplicity that $\inf f = 0$. Set $\varphi:\mathbb R^d\to\mathbb R\cup\{+\infty\}$ as $\varphi(x):=\sqrt{f(x)}$. It is known that \begin{align*} \partial\varphi(x) = \frac{1}{2\sqrt{f(x)}}\partial f(x)\quad \text{for all $x\in\text{dom }\partial f$ with $f(x)\neq 0$.} \end{align*} In particular, this implies that, if $\xi\in\partial\varphi(x)$, then $2\sqrt{f(x)}\,\xi\in\partial f(x)$ for all $\xi\in \mathbb R^d$ and $x\in\text{dom }\partial f$ with $f(x)\neq 0$. This can be rephrased as \begin{align}\label{pty} \text{If $x$ is a minimizer of $\sqrt{f} - \xi$, then it is a minimizer of $f-\big(2\sqrt{f(x)}\big)\xi$.} \end{align} The question is whether this is true for nonconvex functions, say for an $f$ which is only proper and lower semicontinuous.
I have tried the following. If we take a minimizer $x$ of $\sqrt{f}-\xi$, then for $s\in\mathbb R^d$, \begin{equation} \sqrt{f(x)}-\xi x\le \sqrt{f(s)}-\xi s\quad\text{and hence } \sqrt{f(x)} - \sqrt{f(s)}\le \xi(x-s). \end{equation} If we multiply both sides by $\sqrt{f(x)}+\sqrt{f(s)}$, we get \begin{align} f(x) - f(s) \le \big(\sqrt{f(x)} + \sqrt{f(s)} \big)\xi(x-s)\quad\forall s\in\mathbb R^d. \end{align} For $s\in\mathbb R^d$ with $f(s)\le f(x)$, it is easy to see that $\xi(x-s)\ge0$ and hence that \begin{align} f(x) - f(s) \le \big(2\sqrt{f(x)}\big)\xi(x-s), \quad\text{that is}\quad f(x) - \big(2\sqrt{f(x)}\big)\xi x\le f(s) - \big(2\sqrt{f(x)}\big)\xi s. \end{align} For $s$ with $f(s)\ge f(x)$, it seems not so easy to conlude the same. In this case, recalling that $\sqrt{f(x)}-\xi x\le \sqrt{f(s)}-\xi s$, multiplying on both sides by $\sqrt{f(x)}$ leads to \begin{align} f(x) - (\sqrt{f(x)})\xi x \le \sqrt{f(x)}\sqrt{f(s)} - \xi s\le f(s) - (\sqrt{f(x)})\xi s. \end{align} But the $2$ next to $\sqrt{f(x)}$ is missing. Here is where I see the main problem. Is there any reference for these type of questions?
Let us prove the following statement for $g(x) \ge 0, x\in \text{dom } g$, which covers the one in the OP by setting $g(x)=\sqrt{f(x)}$:
As $y$ is a minimizer of $g(x)-a^Tx$, then for any $x \in \text{dom } g $:
$$g(x)-g(y)\ge a^T(x-y) \tag{1}. $$
To prove that $y$ is a minimizer of $g^2(x)-2g(y)a^Tx$, we need to show that for any $x \in \text{dom } g $:
$$g^2(x)-2g(y)a^Tx \ge g^2(y)-2g(y)a^Ty \\ \equiv \color{blue}{(g(x)+g(y))(g(x)-g(y))\ge 2g(y)a^T(x-y)} \tag{2}. $$
This completes the proof.
Following the same method, the following extension can be proven for any $k \in \mathbb N$:
$$\color{blue}{y\in \text{argmin}\left(g(x)-a^Tx \right) \Rightarrow y\in \text{argmin}\left(g^k(x)-kg(y)^{k-1}a^Tx\right)}.$$
It is equivalent to show
$$\left (\sum_{i=0}^{k-1}g(x)^ig(y)^{k-i} \right )(g(x)-g(y))\ge kg(y)^{k-1}a^T(x-y). $$
The above can be further extended as follows for any strictly increasing differentiable convex function $\color{blue}{H: \mathbb R_{\ge 0} \to \mathbb R_{\ge 0}}$:
$$\fbox{$\color{blue}{y\in \text{argmin}\left(g(x)-a^Tx \right) \Rightarrow y\in \text{argmin}\left(H(g(x))-H'(g(y))a^Tx\right)}$}.$$
Using a similar proof method, it is implied from the observation that for any fixed $y\in \mathbb R_{\ge 0}$, the function
$$\frac{H(x)-H(y)}{x-y}$$
is non-negative and increasing in $x$ over both intervals $(0,y)$ and $(y,\infty)$ (note that for the convex function $H$ we have $H(y)-H(x)\ge H'(x)(y-x)$ for any $x,y\in \mathbb R_{\ge 0}$; the limit of the function is $H'(y)$ as $x \to y$).
The problem is special and interesting. The result can be used in a two-person Stackelberg game where the cost functions of the leader and follower are $g(x_L)-a^Tx_L$ and $g^2(x_F)-2g(x_L)a^Tx_F$, respectively. Here, we proved that any point $(y,y)$ with $y \in \text{argmin} (g(x)-a^Tx)$ is a pure-strategy equilibrium of the game.