4.1. Find the maximum of the expression $a / b + b / c$ given $0 < a \leqslant b \leqslant a + b \leqslant c$.
4.1. Solution. For $a=b=1, c=2$ we get the value $3 / 2$. We will prove that this is the maximum. We have

$$
\frac{b}{c}-\frac{1}{2} \leqslant \frac{b}{a+b}-\frac{1}{2}=\frac{b-a}{2(a+b)} \leqslant \frac{b-a}{b}=1-a / b
$$

from which $\frac{b}{c}+\frac{a}{b} \leqslant 3 / 2$.