Problem 1302

7.22. We will consider that the quintet of numbers $\alpha, \beta, \gamma, \delta, \varepsilon$ coincides (up to order) with the quintet of numbers $x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{1} \leqslant x_{1}$. Then

$$\begin{aligned} & \left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\right)\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\frac{1}{x_{4}}+\frac{1}{x_{5}}\right)= \\ & =5+\sum_{i<j}\left(\frac{x_{i}}{x_{j}}+\frac{x_{j}}{x_{i}}\right)=25+\sum_{i<j}\left(\frac{x_{l}}{x_{j}}+\frac{x_{j}}{x_{i}}-2\right) \end{aligned}$$

where the sum is taken over all $C_{6}^{2}=10$ pairs of indices $1 \leqslant i<j \leqslant 5$. Further, we have

$$\left(\sqrt{\frac{p}{q}}-\sqrt{\frac{q}{p}}\right)^{2}=\frac{p}{q}+\frac{q}{p}-2$$

Let $f(x, y)=\frac{x}{y}+\frac{y}{x}-2$, then it is sufficient to prove the inequality

$$\sum_{i<j} f\left(x_{i}, x_{j}\right) \leqslant 6 f(p, q)$$

We will prove that for positive numbers $x \leqslant y \leqslant z$ the inequality $f(x, y)+f(y, z) \leqslant f(x, z)$ holds, i.e.,

$$\frac{x}{y}+\frac{y}{x}-2+\frac{y}{z}+\frac{z}{y}-2<\frac{x}{z}+\frac{z}{x}-2$$

Notice that

$$\frac{x}{y}+\frac{y}{z}-\frac{x}{z}-1=\left(\frac{x}{y}-1\right)\left(1-\frac{y}{z}\right)<0$$

since $x / y \leqslant 1$ and $1 \geqslant y / z$. Similarly, we obtain the estimate

$$\frac{z}{y}+\frac{y}{x}-\frac{z}{x}-1<0$$

which, when added to the previous one, gives the required inequality. Moreover, equality is achieved if and only if either $x=y$ or $y=z$. Using the proven relation, we obtain the following series of inequalities

$$\begin{gathered} f\left(p, x_{1}\right)+f\left(x_{1}, x_{2}\right)+f\left(x_{2}, x_{3}\right)+f\left(x_{3}, x_{4}\right)+f\left(x_{4}, x_{5}\right)+f\left(x_{5}, q\right) \leqslant f(p, q), \\ f\left(p, x_{1}\right)+f\left(x_{1}, x_{3}\right)+f\left(x_{3}, x_{5}\right)+f\left(x_{5}, q\right) \leqslant f(p, q), \\ f\left(p, x_{1}\right)+f\left(x_{1}, x_{4}\right)+f\left(x_{4}, q\right) \leqslant f(p, q), \\ f\left(p, x_{2}\right)+f\left(x_{2}, x_{5}\right)+f\left(x_{5}, q\right) \leqslant f(p, q), \\ f\left(p, x_{2}\right)+f\left(x_{2}, x_{4}\right)+f\left(x_{4}, q\right) \leqslant f(p, q), \\ f\left(p, x_{1}\right)+f\left(x_{1}, x_{5}\right)+f\left(x_{5}, q\right) \leqslant f(p, q) \end{gathered}$$

Adding these and discarding the positive terms of the form $f\left(p, x_{i}\right)$ and $f\left(x_{i}, q\right)$, we obtain the required inequality. Now let us determine when equality holds. Notice that if $p<x_{2}$ or $x_{1}<q$, then equality is not achieved, since then among the discarded terms $f\left(p, x_{2}\right)$ or $f\left(x_{4}, q\right)$ there is a non-zero one. Therefore, for equality, it is necessary that the relations $x_{1}=x_{2}=p$ and $x_{4}=x_{5}=q$ hold. Then from the second of the six inequalities listed above, we get

$$f\left(p, x_{3}\right)+f\left(x_{2}, q\right) \leqslant f(p, q)$$

Since equality must also hold in this inequality, either $p=x_{3}$ or $x_{3}=q$. Transitioning to the original formulation, we obtain that equality is achieved if and only if two numbers from the quintet of numbers $\alpha, \beta, \gamma, \delta, \varepsilon$ coincide with one end of the interval $[p ; q]$, and the other three numbers coincide with the other end of the interval.