Thanks for this great counterexample!

The implied density function is $(2-3t/2)1\{0<t\le 1/3\}+$ $1\{1/3<t\le 2/3\}+$ $3/2(1-t)1\{2/3<t<1\}$ which is non-increasing.

Thanks a lot for your answer! I agree that $X$ and $Y$ are uniformly distributed on [0,1] but I think that there is a mistake in the density of the minimum. For example, consider $\Pr(\min(X,Y)\le 1/3)$. This probability is the probability of 3 dark-brown squares plus the probability of 2 light-brown squares. This gives us 7/12. On the other hand, from your graph this probability is 1/6. In general, I calculated for your example that $\Pr(\min\{X,Y\}\le t)$ equals $2t-3t^2/4$ for $t\in (0, 1/3]$, $1/4+t$ for $t\in(1/3, 2/3]$ and $1/4+3t/2 - 3t^2/4$ for $t\in(2/3, 1)$