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The inequality $-\frac{2}{3}a_n+\frac{1}{3}\le0$ is algebraically equivalent to $\frac{1}{3}\le\frac{2}{3}a_n$, which simplifies to $\frac{1}{2}\le a_n$. Since the lower bound $a_n\ge\frac{1}{2}$ was already proven by induction, the difference $a_{n+1}-a_n$ is guaranteed to be non-positive.

Conditions: The expression for the difference is $a_{n+1}-a_n = -\frac{2}{3}a_n + \frac{1}{3}$.; The lower bound $a_n \ge \frac{1}{2}$ has been established for all $n \ge 1$.

The inequality $-\frac{2}{3}a_n+\frac{1}{3}\le0$ is algebraically equivalent to $\frac{1}{3}\le\frac{2}{3}a_n$, which simplifies to $\frac{1}{2}\le a_n$. Since the lower bound $a_n\ge\frac{1}{2}$ was already proven by induction, the difference $a_{n+1}-a_n$ is guaranteed to be non-positive.

Conditions: The expression for the difference is $a_{n+1}-a_n = -\frac{2}{3}a_n + \frac{1}{3}$.; The lower bound $a_n \ge \frac{1}{2}$ has been established for all $n \ge 1$.