Romanian Mathematical Olympiad

Let $g : [0, 1] \to \mathbb{R}$ defined by $g(x) = x f(x)$, for $x \in [0, 1]$. We have $g(0) = 0 = g(1)$, $g'(x) = f(x) + x f'(x)$, so that:
$$\begin{align*} \int_0^1 x^2 (f'(x))^2 dx &= \int_0^1 (g'(x) - f(x))^2 dx \\ &= \int_0^1 (g'(x))^2 dx - 2 \int_0^1 g'(x) f(x) dx + \int_0^1 (f(x))^2 dx \\ &= \int_0^1 f^2(x) dx + \int_0^1 (g'(x))^2 dx - \\ & \qquad -2(f(1)g(1) - f(0)g(0)) + 2 \int_0^1 g(x) f'(x) dx, \end{align*}$$
implying
$$\begin{align*} \int_{0}^{1} x^2 (f'(x))^2 dx &= \\ &= \int_{0}^{1} f^2(x) dx + \int_{0}^{1} (g'(x))^2 dx + \int_{0}^{1} x \cdot (2f(x)f'(x)) dx \\ &= \int_{0}^{1} f^2(x) dx + \int_{0}^{1} (g'(x))^2 dx + \int_{0}^{1} x \cdot (f^2(x))' dx \\ &= \int_{0}^{1} f^2(x) dx + \int_{0}^{1} (g'(x))^2 dx + (1 \cdot f^2(1) - 0 \cdot f^2(0)) - \int_{0}^{1} f^2(x) dx \\ &= \int_{0}^{1} f^2(x) dx + \int_{0}^{1} (g'(x))^2 dx - \int_{0}^{1} f^2(x) dx = \int_{0}^{1} (g'(x))^2 dx. \end{align*}$$

$$\left( \int_0^1 (2x-1) \cdot g'(x) \, dx \right)^2 \le \int_0^1 (2x-1)^2 \, dx \cdot \int_0^1 (g'(x))^2 \, dx = \\ = \frac{1}{6} \cdot \left( (2 \cdot 1 - 1)^3 - (2 \cdot 0 - 1)^3 \right) \cdot \int_0^1 x^2 (f'(x))^2 \, dx = \frac{1}{3} \cdot \int_0^1 x^2 (f'(x))^2 \, dx.$$
This gives
$$\begin{align*} \int_0^1 (x f'(x))^2 \, dx &\ge 3 \cdot \left( \int_0^1 (2x-1) \cdot g'(x) \, dx \right)^2 = \\ &= 3 \cdot \left( (2 \cdot 1 - 1)g(1) - (2 \cdot 0 - 1)g(0) - 2 \int_0^1 g(x) \, dx \right)^2 = \\ &= 12 \left( \int_0^1 x f(x) \, dx \right)^2, \end{align*}$$
concluding the proof.