For which natural numbers $n \geq 2$ does the inequality $x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2} \geq p\left(x_{1} x_{2}+x_{2} x_{3}+\ldots+x_{n-1} x_{n}\right)$ hold for any real numbers $x_{1}, x_{2}, \ldots, x_{n}$, if

a) $p=1$;

b) $p=4 / 3$;

c) $p=6 / 5$?
Note that if we set several of the last variables $x_{k}, x_{k+1}, \ldots, x_{n}$ to zero in the inequality $x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}-p x_{1} x_{2}-p x_{2} x_{3}-\ldots-p x_{n-1} x_{n} \geq 0 \quad$ (*), we obtain a similar inequality corresponding to a smaller $n$. Hence, if (*) holds (for all $x_{1}, x_{2}, \ldots, x_{n}$) for some $n$, it also holds for smaller $n$. Therefore, for each fixed $p$, there exists a certain integer $N(p) \geq 2$ such that (*) holds for $n < N(p)$ and does not hold (for some $x_{1}, x_{2}, \ldots, x_{n}$) for $n \geq N(p)$; it is also possible that (*) holds for all $n$ (in which case we can consider $N(p)=\infty$).

a) For $p=1$, the inequality (*) is equivalent to the obvious inequality

$x_{1}^{2}+\left(x_{1}-x_{2}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}+\ldots+\left(x_{n-1}-x_{n}\right)^{2}+x_{n}^{2} \geq 0 \ldots$. Therefore, $N(1)=\infty$.

b) For $p=4 / 3$ and $n=3$, the inequality (*) is equivalent to the inequality $\left(3 x_{1}-2 x_{2}\right)^{2}+x_{2}^{2}+\left(2 x_{2}-3 x_{3}\right)^{2} \geq 0$, while for $n=4$, the inequality (*) is violated, for example, when $x_{1}=x_{4}=2, x_{2}=x_{3}=3$.

c) For $p=6 / 5$ and $n=4$, (*) is equivalent to the inequality

$\left(5 x_{1}-3 x_{2}\right)^{2}+x_{2}^{2}+15\left(x_{2}-x_{3}\right)^{2}+x_{3}^{2}+\left(3 x_{3}-5 x_{4}\right)^{2} \geq 0$, while for $n=5$ it is violated, for example, when $x_{1}=x_{5}=9, x_{2}=x_{4}=15, x_{3}=16$.

## Answer

a) For all $n$;
b) for $n \leq 3$;
c) for $n \leq 4$.

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