3. Find the maximum value of the expression

$$
\frac{a}{x}+\frac{a+b}{x+y}+\frac{a+b+c}{x+y+z}
$$

where $a, b, c \in[2,3]$, and the triplet of numbers $x$, $y$, and $z$ is some permutation of the triplet of numbers $a, b, c$.
Answer: 3.75.

Solution. We will prove that

$$
\frac{a}{x}+\frac{a+b}{x+y}+\frac{a+b+c}{x+y+z} \leqslant \frac{15}{4}
$$

Notice that the last fraction equals 1 and $a / x \leqslant 3 / 2$. It remains to show that $\frac{a+b}{x+y} \leqslant \frac{5}{4}$. Since the numbers $x$ and $y$ are some pair of numbers from $a, b$, and $c$, one of them coincides with $a$ or $b$. Thus, it is sufficient to prove that $\frac{u+v}{v+w} \leqslant \frac{5}{4}$ for $u, v, w \in [2,3]$. Or equivalently, $4 u \leqslant v + 5 w$. But this is obvious, since $4 u \leqslant 12 \leqslant v + 5 w$.

If $a=z=3, b=c=x=y=2$, then

$$
\frac{a}{x}+\frac{a+b}{x+y}+\frac{a+b+c}{x+y+z}=\frac{3}{2}+\frac{3+2}{2+2}+\frac{3+2+2}{2+2+3}=\frac{3}{2}+\frac{5}{4}+1=\frac{15}{4}
$$