Determine the narrowest interval into which the product of the sum of three positive numbers and the sum of their reciprocals can fall.
Solution. Let the three positive numbers be $a, b$, and $c$. Then:

$$
\begin{aligned}
(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) & =1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1= \\
& =3+\frac{a}{b}+\frac{b}{a}+\frac{a}{c}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b}
\end{aligned}
$$

Since $a, b, c$ are positive numbers, $\frac{a}{b}+\frac{b}{a} \geq 2, \frac{a}{c}+\frac{c}{a} \geq 2$ and $\frac{b}{c}+\frac{c}{b} \geq 2$, therefore

$$
3+\frac{a}{b}+\frac{b}{a}+\frac{a}{c}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b} \geq 3+2+2+2=9
$$

Thus, it is certain that the value of the expression is at least 9. The value 9 is achieved when $a=b=c=1$.

We will show that the value of the expression can be arbitrarily large, i.e., it is not bounded from above.

Let $K$ be a number greater than 9. We will prove that for suitable $a, b, c$,

$$
(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)>K
$$

Let, for example, $a=\frac{K}{2}, b=\frac{K}{2}, c=1$, then:

$$
\left(\frac{K}{2}+\frac{K}{2}+1\right)\left(\frac{2}{K}+\frac{2}{K}+1\right)>\left(\frac{K}{2}+\frac{K}{2}+1\right) \cdot 1=K+1>K
$$

Since $K$ can be arbitrarily large, the desired narrowest interval is: $[9 ; \infty[$.