# Show $(\int_{-\infty}^\infty \sqrt{p}\sqrt{q}d\mu)^2\leq 2 \int_{-\infty}^\infty \min\{p,q\}d\mu$

Consider a random variable $X$ in $(\Omega, \mathcal{F}, \mathbb{P})$. Let $p,q$ be two densities with respect to a measure $\mu$ in $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ where $\mathcal{B}(\mathbb{R})$ is the Borel $\sigma$-algebra in $\mathbb{R}$. Could you help me to show that $$\left(\int_{-\infty}^\infty \sqrt{p}\sqrt{q}d\mu\right)^2\le 2 \int_{-\infty}^\infty\min\{p,q\}d\mu$$ from van der Vaart "Asymptotic Statistics" proof of Lemma 14.31.

My attempt:

\begin{align}\left(\int_{-\infty}^\infty \sqrt{p}\sqrt{q}d\mu\right)^2&\le \left(\int_{-\infty}^\infty \min\{\sqrt{p},\sqrt{q}\}\left(\sqrt{p}+\sqrt{q}\right)d\mu\right)^2\\[0.2cm]& \le \int_{-\infty}^{\infty}\left(\sqrt{p}+\sqrt{q}\right)^2d\mu \int_{-\infty}^{\infty}\min\left\{\sqrt{p},\sqrt{q}\right\}^2d\mu\\[0.2cm]&=\left(2+2\int_{-\infty}^{\infty}\sqrt{p}\sqrt{q}d\mu\right)\int_{-\infty}^{\infty}\min\{p,q\}d\mu=\dots?\end{align}