THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

a) Observe that $a_n>0$ and that $a_{n+1}-2=\frac{2-a_n}{a_n}$ for $n\ge 1$, so $a_{2n-1}>2>a_{2n}$ for $n\ge 1$. Thus $a_{2n-1}+a_{2n}-4=\frac{a_{2n-1}^2-3a_{2n-1}+2}{a_{2n-1}}>0$ for $n\ge 1$.

$|a_n-2|=\frac{|a_{n-1}-2|}{a_{n-1}}=\cdots =\frac{|a_1-2|}{a_{n-1}\dots a_1}\le \frac{1}{2^{[n/2]}}|a_1-2|\to 0$ shows that $a_n\to 2$.

b) For $x=0$ we get $a<\frac{a_1+\cdots +a_n}{n}$, so $a\le {\lim }_{n\to \infty }\frac{a_1+\cdots +a_n}{n}=2$. We will show that the inequality is true for $a=2$ giving $a_{\max }=2$. It is enough to show that $\sqrt{x^2+a_{2n-1}^2}+\sqrt{x^2+a_{2n}^2}>2\sqrt{x^2+4}$, for any $n\in {\mathbb{N}}^{\ast }$, which is equivalent to $a_{2n-1}^2+a_{2n}^2+2\sqrt{(x^2+a_{2n-1}^2)(x^2+a_{2n}^2)}>2x^2+16$. As $a_{2n-1}+a_{2n}>4$ involves $a_{2n-1}^2+a_{2n}^2>8$, it is sufficient to show that $a_{2n-1}{a}_{2n}\ge 4$. This is an immediate consequence of $a_{2n-1}>2$ and $a_{2n}=1+2/a_{2n-1}$.