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Prove that for any smooth function $\phi:\mathbb{R}\rightarrow\mathbb{R}$ such that $|\frac{d\phi}{dx}|<1$ 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

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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 $$\left|\int_0^\pi e^{i\phi(x)}\ dx \right| > \frac{2}{k} \sin\left(\frac{k \pi}{2}\right)$$

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That solution is brilliant. Although it's clearly logically sound, I can't seem to figure out how you got it. The use of Euler's formula seems obvious in retrospect, but the rest seems to come from nowhere. Would it be possible to provide some idea of what motivated this solution, if only to satisfy my curiosity? –  user26431 Mar 7 '12 at 3:02
    
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

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