Determine the greatest real number $a$ such that the inequality

$$
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2} \geq a\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+x_{4} x_{5}\right)
$$

holds for every five real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$.
Note that

$$
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}=\left(x_{1}^{2}+\frac{x_{2}^{2}}{3}\right)+\left(\frac{2 x_{2}^{2}}{3}+\frac{x_{3}^{2}}{2}\right)+\left(\frac{x_{3}^{2}}{2}+\frac{2 x_{4}^{2}}{3}\right)+\left(\frac{x_{4}^{2}}{3}+x_{5}^{2}\right) .
$$

Now applying the inequality $a^{2}+b^{2} \geq 2 a b$ we get

$$
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2} \geq \frac{2}{\sqrt{3}}\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+x_{4} x_{5}\right) \text {. }
$$

This proves that $a \geq \frac{2}{\sqrt{3}}$. In order to prove $a \leq \frac{2}{\sqrt{3}}$ it is enough to notice that for $\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)=(1, \sqrt{3}, 2, \sqrt{3}, 1)$ we have

$$
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}=\frac{2}{\sqrt{3}}\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+x_{4} x_{5}\right) \text {. }
$$