Let the distance of object from the lens is \(x\)

So the distance of image and lens be \(d - x\)

Using the lens equation :

\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

\(\frac{1}{f}=\frac{1}{d-x}-\frac{1}{-d}\)

\(x =\frac{ d \pm \sqrt{ d ^2-4 fd }}{2}\)

So for real \(x\)

\(\sqrt{ d ^2-4 fd } \geq 0\)

So \(d \geq 4 f\)

So minimum distance between object and image is \(4 f\).