Prove that for any smooth function $\phi:\mathbb{R}\rightarrow\mathbb{R}$ such that $\left|\frac{d\phi}{dx}\right|<1$ for all $x$ in $(0,\pi)$,
$\left(\int_0^\pi \cos(\phi(x)) \; dx\right)^2 + \left(\int_0^\pi \sin(\phi(x))\;dx\right)^2 > 4.$
Prove that for any smooth function $\phi:\mathbb{R}\rightarrow\mathbb{R}$ such that $\left|\frac{d\phi}{dx}\right|<1$ for all $x$ in $(0,\pi)$,
$\left(\int_0^\pi \cos(\phi(x)) \; dx\right)^2 + \left(\int_0^\pi \sin(\phi(x))\;dx\right)^2 > 4.$
Your inequality says $\left| \int_0^\pi e^{i\phi(x)}\ dx \right| > 2$. In fact if |\phi'| < 1 we have $|\phi(x) - \phi(\pi/2)| < |x - \pi/2|$ for $0 < x < \pi$, so $\left| \int_0^\pi e^{i\phi(x)}\ dx \right| \ge \text{Re}\ e^{-i \phi(\pi/2)} \int_0^\pi e^{i \phi(x)}\ dx = \int_0^\pi \cos(\phi(x)-\phi(\pi/2))\ dx > \int_0^\pi \cos(x - \pi/2)\ dx = 2$ Here I'm using the facts that $\cos$ is even and is decreasing on $[0,\pi]$. No smoothness of $\phi$ is necessary, just differentiability.
More generally, if you replaced |\phi'| < 1 by |\phi'| < k where $0 < k < 2$, the inequality would become $\left|\int_0^\pi e^{i\phi(x)}\ dx \right| > \frac{2}{k} \sin\left(\frac{k \pi}{2}\right)$