A sum above one, a product below 3/16

Idea. The probability is the area of the region above the line $y=1-x$ and below the hyperbola $y=\frac{3}{16x}$, inside the square. The two curves cross at $x=\frac{1}{4}$ and $x=\frac{3}{4}$, so the region is two thin pieces near the corners $(0,1)$ and $(1,0)$.

Step 1: the crossings. $1-x=\frac{3}{16x}$ gives $16x^2-16x+3=0$, so $x=\frac{1}{4}$ or $x=\frac{3}{4}$. For $\frac{1}{4}<x<\frac{3}{4}$ the hyperbola is below the line and there is no region.

Step 2: the piece near $(0,1)$. For $x\le \frac{3}{16}$ the hyperbola is above the top of the square, so $y$ runs from $1-x$ to 1: area ${\int }_0^{3/16}xdx=\frac{9}{512}$. For $\frac{3}{16}\le x\le \frac{1}{4}$, $y$ runs from $1-x$ to $\frac{3}{16x}$: $$\begin{array}{rl}& {\int }_{3/16}^{1/4}\left(\frac{3}{16x}-1+x\right)dx\\ & =\frac{3}{16}\ln \frac{4}{3}-\frac{1}{16}+\frac{7}{512}.\end{array}$$

Step 3: the piece near $(1,0)$. For $\frac{3}{4}\le x\le 1$: $$\begin{array}{rl}& {\int }_{3/4}^1\left(\frac{3}{16x}-1+x\right)dx\\ & =\frac{3}{16}\ln \frac{4}{3}-\frac{1}{4}+\frac{7}{32}.\end{array}$$

Step 4: add up. $$\begin{array}{rl}& \frac{3}{8}\ln \frac{4}{3}+\frac{9}{512}+\frac{7}{512}-\frac{1}{16}-\frac{1}{4}+\frac{7}{32}\\ & =\frac{3}{8}\ln \frac{4}{3}-\frac{1}{16}\approx \mathrm{0.0454.}\end{array}$$

A check. Swapping $x$ and $y$ maps one piece onto the other, and indeed both pieces have area $\frac{3}{16}\ln \frac{4}{3}-\frac{1}{32}$.