Why is the integral of a square always larger than the square of an integral?

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I learned in physics that $\langle x^2 \rangle - \langle x \rangle ^2 = \sigma_x^2 \ge 0$ and thus $\langle x^2 \rangle \ge \langle x \rangle ^2$. In the case of continuous distribution, it becomes

$$\int_a^b \! [f(x)]^2 \, \mathrm{d}x \ge \left(\int_a^b \! f(x) \, \mathrm{d}x \right)^2.$$

Does the inequality always hold? Is there a proof of the inequality?

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The inequality indeed holds more generally. Have a look at Jensen's inequality, Cauch-Schwarz's inequality, and Holder's inequality on Wikipedia. These are all generalizations of the inequality that you have asked about.

Proofs of these inequalities can be found in books on functional analysis, see for example Walter Rudin - Real and Complex Analysis.