The Problems of Ukrainian Authors

Answer: $\gamma \ge n$.

At first we will show that for $0<\gamma <n$ there exists a set $x_0,x_1,\dots ,x_n$, for which the inequality from the statement of the problem is not held.

Let $x_0=x^{-n}$, $x_1=x_2=\cdots =x_n=x$ for some $x>0$. Then we have $$\begin{array}{c}{x}_0^{\gamma }+x_1^{\gamma }+\cdots +x_n^{\gamma }=x^{-n\gamma }+n{x}^{\gamma },\\ \frac{1}{x_0}+\frac{1}{x_1}+\cdots +\frac{1}{x_n}=x^n+n{x}^{-1}.\end{array}$$ For $\gamma <n$ we want to find $x>0$ for which $x^n+n{x}^{-1}>x^{-n\gamma }+n{x}^{\gamma }$. For this purpose it is enough to choose $x>1$: $x^n+n{x}^{-1}>x^n>(n+1)x^{\gamma }>n{x}^{\gamma }+x^{-n\gamma }$, i.e. $x^{-n\gamma }>n+1$ also $x>\sqrt[n]{n+1}$. From here we find a corresponding example.

Let now $\gamma =n$ then by the Cauchy's inequality we receive $$\begin{array}{c}\frac{x_0^n+x_1^n+\cdots +x_{n-1}^n}{n}\ge x_0{x}_1\dots x_{n-1}=\frac{1}{x_n},\\ \frac{x_0^n+x_1^n+\cdots +x_{n-2}^n+x_n^n}{n}\ge x_0{x}_1\dots x_{n-2}{x}_n=\frac{1}{x_{n-1}},\dots ,\end{array}$$ $$\frac{x_1^n+x_2^n+\cdots +x_n^n}{n}\ge x_1{x}_2\dots x_n=\frac{1}{x_0}.$$ If all these inequalities are added then we will receive the required inequality.

Now we consider the case $\gamma >n$, if we prove that for any set of positive numbers $x_0,x_1,\dots ,x_n$ such that $x_0{x}_1\dots x_n=1$ the inequality $x_0^{\gamma }+x_1^{\gamma }+\cdots +x_n^{\gamma }\ge x_0^n+x_1^n+\cdots +x_n^n$ holds, then we will receive a necessary inequality for $\gamma$ if we use the already proved inequality for $n$: $$\begin{gathered} x_0^{\gamma }+\cdots +x_n^{\gamma }\ge x_0^n+\cdots +x_n^n\iff \\ {x}_0^{\gamma }+\cdots +x_n^{\gamma }\ge x_0^{n+\frac{\gamma -n}{n+1}}{x}_1^{\frac{\gamma -n}{n+1}}\dots x_n^{\frac{\gamma -n}{n+1}}+\cdots +x_0^{\frac{\gamma -n}{n+1}}\dots x_{n-1}^{\frac{\gamma -n}{n+1}}{x}_n^{n+\frac{\gamma -n}{n+1}}. \end{gathered}$$ To prove the last inequality we will take advantage of a weighted Cauchy's inequality $$x_0^{n+\frac{\gamma -n}{n+1}}{x}_1^{\frac{\gamma -n}{n+1}}\dots x_n^{\frac{\gamma -n}{n+1}}\le \frac{n+\frac{\gamma -n}{n+1}}{\gamma }{x}_0^{\gamma }+\frac{\gamma -n}{\gamma (n+1)}{x}_1^{\gamma }+\cdots +\frac{\gamma -n}{\gamma (n+1)}{x}_n^{\gamma }.$$ We will write down similar inequalities for every item, we will add them and we will receive the requisite evidence.