Geometric interpretation of an integral inequality Let $f: [a, b] \to \mathbb [0, \infty)$ be an integrable function.
By the Cauchy-Schwarz inequality:
$$ \left(\int_a^b f(x) dx\right)^2 \leq (b-a) \int_a^b f(x)^2 dx$$
with equality iff $f$ is constant.
If we notate $\mu(f) := \frac{1}{b-a} \int_a^b f$, dividing with $\frac{1}{(b-a)^2}$ we get 
$$\mu(f)^2 \leq \mu(f^2)$$
i.e. the square of the mean of a function is no larger than the mean of the square of a function, and the equality occurs iff $f$ is constant.
Is there an elegant intuitive way to see this geometrically? The best I can do is to use a discrete analogue while approximating $f$ and $f^2$ with rectangles (the definition of the integral).
Is there a more visual approach?
 A: Here's one way to look at it. I don't know how intuitive it is, because I can't find a good way to express the geometric reasoning, but maybe it says something to someone.
If the curve $f$ between $a$ and $b$ rotates around the $x$-axis, the resulting object $A$ has volume
$$
V_A = \pi \int_a^b f(x)^2\,\mathrm{d}x = \pi (b-a) \mu(f^2).
$$
Let $g=\mu(f)$, the average of $f$. If the curge $g$ rotates around the $x$ axis, the resulting object $B$ is a cylinder with volume
$$
V_B = \pi (b-a) \mu(f)^2.
$$
Now compare $A\setminus B$, that is, the part of $A$ which is outside $B$, with $B\setminus A$, the part of $B$ which is outside $A$. Your result is equivalent to the fact that the volume of $A\setminus B$ is larger than the volume of $B \setminus A$. 
Unfortunately I can't think of a good way to explain this geometrically, but it feels intuitive, if you remember that the radius of $B$ is the average of the radius of $A$. My intuitive idea is similar to comparing three circles of radii $r-r'$, $r$, and $r+r'$: clearly the difference between the areas of the two smallest circles is smaller than the difference between the areas of the two largest circles.
A: Combining JiK's (and implicitly john mangual's) answer, and Semiclassical's comments: If the region enclosed by the $x$-axis and the graph $y = f(x) - \mu(f)$ (subject to $a \leq x \leq b$) is revolved about the $x$-axis, the (non-negative) volume swept out (multiplied by $1/\pi$) is given by
\begin{align*}
\int_{a}^{b} \bigl[f(x) - \mu(f)\bigr]^{2}\, dx
  &= \int_{a}^{b} \bigl[f(x)^{2} - 2f(x)\mu(f) + \mu(f)^{2}\bigr]\, dx \\
  &= \left[\int_{a}^{b} f(x)^{2}\, dx\right] - \mu(f)^{2}.
\end{align*}

The same volume is swept out by revolving the graph $y = \bigl|f(x) - \mu(f)\bigr|$, if you want to avoid revolving a region that dips below the $x$-axis.
A: In physics this is known as the parallel axis theorem.  

The Moment of intertia of a body $A \in \mathbb{R}^2$ is the average distance squared relative to the center of Mass:
$$ I_{CM} = \mathbb{E}\big[ || r - \overline{r}||^2\big] = \int_A \big[(x-x_0)^2 + (y-y_0)^2\big] \;dA$$
How do you compute moments relative to other axes?  Then we have formula:
$$ I = I_{CM} + m d^2 $$
$m$ is the mass of your object and $d$ is the distance between the center of your two axes.  In probability theory we could say:
$$ \mathbb{E}[ (X - a)^2] =  \mathbb{E}[ X^2] -  2a  \mathbb{E}[ X] +  a^2 \geq 0 $$
So this is certainly minimized when $a = \mathbb{E}[X]$
