Simple but difficult to prove inequality involving integrals

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 $(0,\pi)$,

$$\left(\int_0^\pi \cos(\phi(x)) \; dx\right)^2 + \left(\int_0^\pi \sin(\phi(x))\;dx\right)^2 > 4.$$

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Just a terminological quibble: an identity says something is equal to something else. This is an inequality. –  Robert Israel Mar 7 '12 at 0:22
@RobertIsrael You're right... apologies. –  user26431 Mar 7 '12 at 0:23

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 $$\left|\int_0^\pi e^{i\phi(x)}\ dx \right| > \frac{2}{k} \sin\left(\frac{k \pi}{2}\right)$$
Well, one ingredient was checking the boundary case $\phi' = 1$ and seeing that you got equality there. That's useful because it means you can't use any estimates that would not be tight in that case. The fact that $|z| = \text{Re}( e^{i\theta} z)$ for suitable $\theta$ is pretty standard. –  Robert Israel Mar 7 '12 at 7:39