Convergence of positive sequence in which each term is less than the average of preceding 2 terms

Yes, such a sequence is necessarily convergent. One possible approach is to show that $(\max(x_n, x_{n+1}))_n$ is a monotonically decreasing sequence which is bounded below and therefore is convergent. Then one can show that $(x_n)$ converges to the same limit.

In fact this works for the more general case where each sequence element is less than (or equal to) the average of the preceding $k$ elements, with a fixed $k \ge 2$. We have the following:

Let $(x_n)$ be a sequence of non-negative real numbers and $k \ge 2$ such that $$ x_{n+k} \le \frac{x_n + \ldots + x_{n+k-1}}{k} $$ for all $n$. Then $(x_n)$ is convergent.

Proof: We consider the sequence $(y_n)$ defined as $$ y_n = \max(x_n, x_{n+1}, \ldots, x_{n+k-1}) \, . $$ Then $x_{n+k} \le y_n$, so that $$ y_{n+1} = \max(x_{n+1}, \ldots, x_{n+k-1}, x_{n+k}) \\ \le \max(x_{n+1}, \ldots, x_{n+k-1}, y_n) = y_n \, , $$ i.e. $(y_n)$ is a decreasing sequence (and bounded below by zero). It follows that $$ a = \lim_{n \to \infty} y_n $$ exists. Now we show that $(x_n)$ has the same limit $a$:

For $\epsilon > 0$ there is an $N$ such that $$ \tag{*} a \le y_n = \max(x_n, x_{n+1}, \ldots, x_{n+k-1}) < a + \epsilon $$ for $n \ge N$. We'll show that $$ a - (k-1)\epsilon < x_n < a + \epsilon $$ for $n \ge N$. The right inequality is clear, since $x_n \le y_n < a + \epsilon$. It remains to show that $a - (k-1)\epsilon < x_n$.

Case 1: $x_n = \max(x_n, x_{n+1}, \ldots, x_{n+k-1})$. Then $(*)$ gives $$ a \le y_n = x_n \, . $$

Case 2: $x_j = \max(x_n, x_{n+1}, \ldots, x_{n+k-1})$ for some $j \in \{ n+1, \ldots n+k-1 \}$. Then $(*)$ gives $$ a \le y_n = x_j \, . $$ It follows that $$ ka \le k x_j \le x_{j-1} + x_{j-2} + \ldots + x_{j-k+1} \le x_n + (k-1) (a + \epsilon) $$ which implies $$ a - (k-1) \epsilon < x_n \, . $$

This concludes the proof.