Given $f$ on an interval give size of integral of $f(x)/x$ (Putnam problem, December 6 2014) Suppose that $f$ is a function on the interval $[1,3]$ such that $f(x)$ is between $-1$ and $1$ for all $x$ and the integral of $ f(x) $ on that interval is 0 . 
What's the largest the integral of $\frac{f(x)}x$ on that interval can be? 

The optimal choice of $f(x)$ is such that its highest values are divided by the lowest values of $x$, i.e. it should have the highest values for the lowest $x$, the highest values need to be "moved left".
Therefore, obviously the optimal choice is: 
$f(x)=1$ for $x\in[1,2]$ and $f(x)=-1$ for $x\in[2,3]$.
The integral is then:
\begin{equation}
\int_1^3\frac{f(x)}xdx=\int_1^2\frac{dx}x-\int_2^3\frac{dx}x
=\ln\left(\frac21\right)-\ln\left(\frac32\right) = \ln\left(\frac43\right)
\end{equation}
