We know:

Let the seperation between object and image by $x$,
$$
x=u-v
$$
Using equation (1), on rewriting,
$$
\begin{aligned}
& x=u-\frac{x f}{u} \\
& \Rightarrow x=\frac{u^2}{(u+f)}---(1)
\end{aligned}
$$
we first take log of equation (1)
$$
\Rightarrow \ln (x)=2 \ln (u)-\ln (u+f)
$$
Subsequently, to minimize $x$ w.r.t. $u$, take the derivative of $x$ w.r.t. $u$ :
$$
\begin{aligned}
& \Rightarrow \frac{1}{x} \frac{d x}{d u}=\frac{2}{u}-\frac{1}{(u+f)} \\
& \Rightarrow \frac{d x}{d u}=\frac{x(u+2 f)}{u(u+f)}---(2)
\end{aligned}
$$
Condition for minimum $x$ requires, $\frac{d x}{d u}=0$ and $\frac{d^2 x}{\partial u^2}<0$
$$
\Rightarrow \frac{d x}{d u}=\frac{x(u+2 f)}{u(u+f)}=0
$$
So, $x=0$ \{is the trivial solution $\}$ and the real solution occurs for $u=-2 f=-R$. Lets check the minimum value using equation (1),
$$
\Rightarrow x_{\min }=\frac{u^2}{(u+f)}=\frac{(-2 f)^2}{(-2 f+f)}=4 f
$$