Problem 1270

Solution. Let $S=a_{1}+a_{2}+\ldots+a_{n}$. Dividing the inequality to be proven by $n$ and taking the square root:

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On the left side, we have the quadratic mean of $n$ positive numbers, which is not greater than the arithmetic mean of these same numbers:

$$\begin{aligned} & \sqrt{\frac{\left(\frac{a_{1}}{a_{2}+\ldots+a_{n}}\right)^{2}+\left(\frac{a_{2}}{a_{3}+\ldots+a_{1}}\right)^{2}+\ldots\left(\frac{a_{n}}{a_{1}+\ldots+a_{n-1}}\right)^{2}}{n}} \geq \\ & \geq \frac{\left(\frac{a_{1}}{a_{2}+\ldots+a_{n}}\right)+\left(\frac{a_{2}}{a_{3}+\ldots+a_{1}}\right)+\ldots+\left(\frac{a_{n}}{a_{1}+\ldots+a_{n-1}}\right)}{n}= \\ & =\frac{\left(\frac{S-\left(a_{2}+\ldots+a_{n}\right)}{a_{2}+\ldots+a_{n}}\right)+\left(\frac{S-\left(a_{3}+\ldots+a_{1}\right)}{a_{3}+\ldots+a_{1}}\right)+\ldots+\left(\frac{S-\left(a_{1}+\ldots+a_{n-1}\right)}{a_{1}+\ldots+a_{n-1}}\right)}{n}= \\ & =\frac{\left(\frac{S}{a_{2}+\ldots+a_{n}}-1\right)+\left(\frac{S}{a_{3}+\ldots+a_{1}}-1\right)+\ldots\left(\frac{S}{a_{1}+\ldots+a_{n-1}}-1\right)}{n}= \\ & =\frac{S \cdot\left(\frac{1}{a_{2}+\ldots+a_{n}}+\frac{1}{a_{3}+\ldots+a_{1}}+\ldots+\frac{1}{a_{1}+\ldots+a_{n-1}}\right)-n}{n}= \\ & =S \cdot \frac{\frac{1}{a_{2}+\ldots+a_{n}}+\frac{1}{a_{3}+\ldots+a_{1}}+\ldots+\frac{1}{a_{1}+\ldots+a_{n-1}}}{n}-1=S \cdot A-1 \end{aligned}$$

where $A$ denotes the resulting complex fraction. Based on the inequality between the arithmetic and harmonic means:

$$\begin{aligned} \frac{1}{A} & =\frac{1}{\frac{1}{a_{2}+\ldots+a_{n}}+\frac{1}{a_{3}+\ldots+a_{1}}+\ldots+\frac{1}{a_{1}+\ldots+a_{n-1}}} \leq \\ & \leq \frac{\left(a_{2}+\ldots+a_{n}\right)+\left(a_{3}+\ldots+a_{1}\right)+\ldots+\left(a_{1}+\ldots+a_{n-1}\right)}{n}=\frac{S \cdot(n-1)}{n} \end{aligned}$$

Thus, $A \geq \frac{n}{S \cdot(n-1)}$, substituting this back in:

$$\begin{aligned} & \sqrt{\frac{\left(\frac{a_{1}}{a_{2}+\ldots+a_{n}}\right)^{2}+\left(\frac{a_{2}}{a_{3}+\ldots+a_{1}}\right)^{2}+\ldots+\left(\frac{a_{n}}{a_{1}+\ldots+a_{n-1}}\right)^{2}}{n}} \geq \\ & \geq S \cdot \frac{n}{S \cdot(n-1)}-1=\frac{n-(n-1)}{n-1}=\frac{1}{n-1} . \end{aligned}$$

This completes the proof of the statement.