Textbook problem:
Two numbers, first $x$ and then $y$, are chosen at random between $1$ and $2$. What is the average value of the quotient $\frac{x}{y}$? Can you argue on elementary grounds that the answer must exceed 1?
My answer: If $x$ has been picked, then the average of the quotient is
$$
\int_1^2 \frac{x}{y}\,dy = x\ln(2)
$$
Averaging over possible values of $x$ yields
$$
\int_1^2 x\ln(2)\, dx = \frac{3}{2}\ln(2) \approx 1.03
$$
But this is not an argument based on elementary grounds I don't think.
Interpreting the average value of a function over an interval as being the height of the rectangle with base on the interval and with the same area as the area under the function over the interval leads me to the following argument.
The average value of $x$ is $3/2$. The quotient $x/y$ then ranges from $3/4$ to $3/2$. We can underestimate the area under this curve by splitting it up into a rectangle with length $1$ and height $3/4$ and a right triangle on top of the rectangle whose hypotenuse is the tangent line to the curve $\frac{3}{2y}$ at the point $(3/2,1)$. This gives a triangle with height $7/12$ and base $7/8$.
So an underestimate for the area under the curve would be $\left(1\cdot \frac{3}{4}\right) + \left(\frac{1}{2}\cdot \frac{7}{8}\cdot \frac{7}{12}\right) = \frac{193}{192}$.
Here is an illustration of my quotient function with $x = 3/2$, the tangent line, and the rectangle:
Since the base of the rectangle with the same area on $[1,2]$ has length $1$ its height must be at least $\frac{193}{192}$ to match the area under the graph. Hence, the height, or average value, exceeds $1$.
Question: Does this seem to be an argument the author could be looking for instead of integrating?
This is a single variable calculus text. The section is ''The Average Value of a Function". Note that not until the next chapter is probability introduced.
Update: I have obtained a copy of the author's own solutions manual. Here is the official solution:
To argue on elementary grounds that the answer must exceed 1, note that the outcomes $a/b$ and $b/a$ must occur equally often, so that the answer is certainly greater than the minimum over all $a$ and $b$ in $[1,2]$ of the average of these two numbers. Now this average is
$$
\frac{1}{2}\left[\frac{a}{b} + \frac{b}{a}\right] = \frac{a^2 + b^2}{2ab}
$$
and this is always greater than 1 because $a^2 + b^2 – 2ab = (a-b)^2 > 0$.
Best Answer
We can argue as follows. The probability distribution is symmetric w.r.t. interchanging $x$ and $y$. This means that we can compute the expectation value by restricting $x$ to be larger than or equal to $y$ and computing the expectation value of $f(x,y) = \frac{1}{2}\left(\frac{x}{y} + \frac{y}{x}\right)$ over this modified probability distribution.
The fact that the expectation value exceeds $1$ then follows from the fact that the function $f(x,y)$ is larger than or equal to $1$:
$$f(x,y) - 1 = \frac{1}{2}\left(\frac{x}{y} + \frac{y}{x}-2\right) = \frac{x^2+y^2-2xy}{2xy} = \frac{(x-y)^2}{2 xy}\geq 0$$
The function $f(x,y)$ is then equal to $1$ when $x = y$ which occurs on a subset of measure zero, so the expectation value is clearly larger than $1$.