We begin with a lemma. Lemma: Let $x, y > 1$ be real numbers. Then $$\frac{1}{x^2-1} + \frac{1}{y^2-1} \geq \frac{2}{xy-1},$$and equality holds only when $x=y$. Proof: Note that $(xy+1)(x-y)^2 \geq 0$, with equality holding only when $x = y$ (since $x, y > 1$, so $xy+1\neq 0$). Therefore, expanding this out, we get that $x^3y+xy^3 - 2x^2y^2 - 2xy + x^2+y^2 \geq 0$, with equality holding only when $x = y$. This implies $x^3y+xy^3 - 2xy - x^2-y^2+ 2\geq 2x^2y^2-2x^2-2y^2+2$, with equality holding only when $x=y$, which is equivalent to $(x^2+y^2-2)(xy-1) \geq 2(x^2-1)(y^2-1)$, with equality only holding when $x=y$. Since $x, y > 1$, we can safely say that this is equivalent to $\frac{x^2+y^2-2}{(x^2-1)(y^2-1)}\geq \frac{2}{xy-1}$, with equality only holding when $x, y > 1$. Thus, $\frac{1}{x^2-1} + \frac{1}{y^2-1} \geq \frac{2}{xy-1}$, with equality holding only when $x=y$, as desired. $\square$ Now, we are ready to tackle both portions. (a): Note that all the numbers on the board will be $>1$ at any time, since initially all of the numbers are $>1$, and if $a,b > 1$, then $\sqrt{\frac{ab+1}{2}} > 1$. Therefore, by the lemma, it readily follows that $$\sum_{s\in \text{board }} \frac{1}{s^2-1}$$is a decreasing monovariant upon each term, since $$\frac{1}{a^2-1} + \frac{1}{b^2-1} \leq \frac{2}{ab-1} = \frac{1}{\left(\sqrt{\frac{ab+1}{2}}\right)^2-1},$$and furthermore equality only holds when we apply the operation on two of the same number. Therefore, it follows that $$\frac{1}{x^2-1} \leq \frac{1}{2^2-1} \cdot n = \frac{n}{3},$$so since $x > 1$ (as all numbers on the board will be $>1$ as discussed earlier), $$x^2-1 \geq \frac{3}{n},$$and so $$x\geq \sqrt{\frac{n+3}{n}},$$as $x > 0$, so we're done with part $(a)$. $\square$ (b): As discussed in (a), the monovariant only stays the same after a move if the two numbers we apply it on are equal. If $n$ is a power of $2$, then we can always make equality hold, since we have $2^k$ twos, which we can then turn into $2^{k-1}$ $\sqrt{\frac{5}{2}}$'s by applying the operation on pairs of twos, which we can then turn into $2^{k-2}$ $\sqrt{\frac{7}{4}}$'s by applying the operation on pairs of $\sqrt{\frac{5}{2}}$'s, etc. Therefore, since equality always holds since each operation is on two of the same number, we have that $$\sum_{s\in \text{board}}\frac{1}{s^2-1}$$is invariant, so we get that $\frac{1}{x^2-1} = \frac{1}{2^2-1}\cdot n$, which implies $x = \sqrt{\frac{n+3}{n}}$ as $x > 0$. Now, if $n$ is odd, then in fact we must apply an operation on two different numbers at least once, since we have an odd number of initial $2$'s, and we cannot create any more $2$s, since all the numbers on the board will be $\leq 2$ at any time, since initially all of the numbers are $\leq 2$, and if $a,b \leq 2$, then $\sqrt{\frac{ab+1}{2}} \leq \sqrt{\frac{5}{2}} < 2$, and furthermore any numbers that are "made" (i.e. not in the original set of $n$ twos) from some $a$ and $b$ on the board, would be at most $\sqrt{\frac{ab+1}{2}} \leq \sqrt{\frac{5}{2}} < 2$, since $a, b\leq 2$. So, since $n$ is odd, we cannot always pair the $2$s to make a $\sqrt{\frac{5}{2}}$, since that would imply that the $n$ twos can be partitioned into a number of pairs, absurd since $n$ is odd. So, equality is not held in one of the moves, so the monovariant $\sum_{s\in \text{board}}\frac{1}{s^2-1}$ strictly decreases from its original value $\frac{n}{3}$, so $\frac{1}{x^2-1} < \frac{n}{3}$, implying $x > \sqrt{\frac{n+3}{n}}$, i.e. the inequality is strict. Therefore, for $n$ a power of $2$, equality can be achieved, and for $n$ odd, the inequality must be strict, and there are infinitely many powers of $2$ and odd numbers, so we're done. $\blacksquare$