Problem 10.8. Non-negative numbers $a, b, c$ sum up to 1. Find the maximum possible value of the quantity

$$
(a+3 b+5 c) \cdot\left(a+\frac{b}{3}+\frac{c}{5}\right)
$$
Answer: $\frac{9}{5}$.

Solution. It is not hard to verify that when $a=\frac{1}{2}, b=0, c=\frac{1}{2}$, the value $\frac{9}{5}$ is obtained. We will prove that it does not exceed $\frac{9}{5}$ for any non-negative $a, b, c$ with a sum of 1.

Notice that

$$
(a+3 b+5 c) \cdot\left(a+\frac{b}{3}+\frac{c}{5}\right)-\frac{9}{5}(a+b+c)^{2}=-\frac{4}{15}\left(3(a-c)^{2}+a b+3 b^{2}+5 b c\right) \leqslant 0
$$

therefore

$$
(a+3 b+5 c) \cdot\left(a+\frac{b}{3}+\frac{c}{5}\right) \leqslant \frac{9}{5}(a+b+c)^{2}=\frac{9}{5}
$$