The functions $f$ and $g$ vanish at no point in the half-open interval $(0, 1]$, so $\left|\int_{0}^{1} f(x) \, dx\right| = \int_{0}^{1} |f(x)| \, dx$ and $\left|\int_{0}^{1} g(x) \, dx\right| = \int_{0}^{1} |g(x)| \, dx$. Consequently,
$$\begin{aligned} 1 &= \int_{0}^{1} 2x \, dx \le \int_{0}^{1} \sqrt{f(x)g(x)} \, dx = \int_{0}^{1} \sqrt{|f(x)||g(x)|} \, dx \\ &\le \int_{0}^{1} \frac{|f(x)| + |g(x)|}{2} \, dx = \frac{1}{2} \left( \int_{0}^{1} |f(x)| \, dx + \int_{0}^{1} |g(x)| \, dx \right), \end{aligned}$$
and the conclusion follows.