10.8. Given a number $a \in(0,1)$. Positive numbers $x_{0}, x_{1}, \ldots, x_{n}$ satisfy the conditions $x_{0}+x_{1}+\ldots+x_{n}=n+a$ and $\frac{1}{x_{0}}+\frac{1}{x_{1}}+$ $+\ldots+\frac{1}{x_{n}}=n+\frac{1}{a}$. Find the minimum value of the expression $x_{0}^{2}+x_{1}^{2}+\ldots+x_{n}^{2}$.

(A. Khryabrov)
Answer: $n+a^{2}$.

Solution: Note that equality is achieved when $x_{0}=a$ and $x_{1}=\ldots=x_{n}=1$.

Write $\sum x_{k}^{2}=\sum\left(1-x_{k}\right)^{2}+2 \sum x_{k}-(n+1)=\sum\left(1-x_{k}\right)^{2}+$
$+n-1+2 a$. It is sufficient to prove that $\sum\left(1-x_{k}\right)^{2} \geqslant(1-a)^{2}$. Let $x_{0}$ be the smallest of the numbers.

If $x_{0} \leqslant a$, we have $\sum\left(1-x_{k}\right)^{2} \geqslant\left(1-x_{0}\right)^{2} \geqslant(1-a)^{2}$.

If $x_{0} \geqslant a$, then $\sum\left(1-x_{k}\right)^{2}=\sum x_{k}\left(\frac{1}{x_{k}}-2+x_{k}\right)$, which, since the expressions in parentheses are non-negative, is not less than $a \sum\left(\frac{1}{x_{k}}-2+x_{k}\right)=a\left(n+\frac{1}{a}-2(n+1)+n+a\right)=$ $=a\left(\frac{1}{a}-2+a\right)=(a-1)^{2}$.

Comment: It is proven that each variable is not less than $a-1$ point.

Answer with an example without justification - 0 points.

An incomplete solution based on the method of Lagrange multipliers (the existence of a minimum point is not mentioned and/or the minimum point is not a boundary point) - 0 points.

An incomplete solution based on the $p q r$-method. For example, it is not strictly proven that when the ratio $q / r$ is fixed, some two roots coincide. Typically, "proofs" are provided that substitute statements for equivalent ones (e.g., "there were other solutions here, but not here, so some two coincide"), or non-formalizable arguments about the movement of graphs (e.g., "the graph moves continuously, so the roots coincide") - 0 points.