We have an inequality similar to Hardy's inequality:
$$
\boxed{\left\|\frac{f(x)}{x}\right\|_{L^2[1,\infty)} ≤ f(1) + 2 \left\|f'\right\|_{L^2[1,\infty)}}
$$
Proof: First remark that
$$
∫_1^\infty \left|\frac{f(x)}{x}\right|^2\mathrm d x = \left\|\frac{1}{x} \left(f(1)+ \int_1^x f'\right)\right\|_{L^2[1,\infty)}^2
\\
≤ \left(\left\|\frac{f(1)}{x}\right\|_{L^2[1,\infty)} + \left\|\frac{1}{x} \int_1^x f'\right\|_{L^2[1,\infty)}\right)^2
$$
and the first integral is easily bounded since
$$
\left\|\frac{f(1)}{x}\right\|_{L^2[1,\infty)} = f(1) \left(\int_1^\infty x^{-2}\,\mathrm d x\right)^{1/2} = f(1)
$$
To bound the second integral, we can do the same strategy as for the classical Hardy's inequality and use first a change of variable $t= sx$ to get
$$
\left\|\frac{1}{x} \int_1^x f'(t)\,\mathrm d t\right\|_{L^2[1,\infty)}
= \left\|\int_{1/x}^1 f'(sx)\,\mathrm d s\right\|_{L^2[1,\infty)}
\\
= \left\|\int_0^1 \mathbf{1}_{\{s>1/x\}} f'(sx)\,\mathrm d s\right\|_{L^2_x[1,\infty)}
\\
≤ \int_0^1 \left\| \mathbf{1}_{\{sx>1\}} f'(sx)\right\|_{L^2_x[1,\infty)} \,\mathrm d s
$$
and then a second change of variable giving
$$
\int_0^1 \left\| \mathbf{1}_{\{sx>1\}} f'(sx)\right\|_{L^2_x[1,\infty)} \,\mathrm d s = \int_0^1 \left(\int_1^\infty |\mathbf{1}_{\{sx>1\}} f'(sx)|^2\,\mathrm d x\right)^{1/2} \,\mathrm d s
\\
= \int_0^1 \left(\int_{s}^\infty |\mathbf{1}_{\{y>1\}} f'(y)|^2\,\,\mathrm d y\right)^{1/2} s^{-1/2}\,\mathrm d s
\\
= \int_0^1 \left(\int_{1}^\infty |f'(y)|^2\,\,\mathrm d y\right)^{1/2} s^{-1/2}\,\mathrm d s = 2 \|f'\|_{L^2[1,\infty)}
$$
Therefore
$$
∫_1^\infty \left|\frac{f(x)}{x}\right|^2\mathrm d x ≤ \left(f(1) + 2 \left\|f'\right\|_{L^2[1,\infty)}\right)^2
$$
