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The example proves the sequence is non-increasing. The mathematical derivation shows $a_{n+1}-a_n \le 0$, which implies $a_{n+1} \le a_n$.

Conditions: The sequence is defined by $a_{n+1}=\frac{1}{3}(a_n+1)$ with $a_1=1$.; The lower bound $a_n \ge \frac{1}{2}$ has been proven.

The example proves the sequence is non-increasing. The mathematical derivation shows $a_{n+1}-a_n \le 0$, which implies $a_{n+1} \le a_n$.

Conditions: The sequence is defined by $a_{n+1}=\frac{1}{3}(a_n+1)$ with $a_1=1$.; The lower bound $a_n \ge \frac{1}{2}$ has been proven.

Subtract $a_n$ from both sides of the recurrence relation. Distribute the $\frac{1}{3}$ and combine like terms to simplify the expression to $-\frac{2}{3}a_n + \frac{1}{3}$.

Conditions: The recurrence relation is $a_{n+1}=\frac{1}{3}(a_n+1)$.; The goal is to analyze the sign of the difference to determine monotonicity.

Subtract $a_n$ from both sides of the recurrence relation. Distribute the $\frac{1}{3}$ and combine like terms to simplify the expression to $-\frac{2}{3}a_n + \frac{1}{3}$.

Conditions: The recurrence relation is $a_{n+1}=\frac{1}{3}(a_n+1)$.; The goal is to analyze the sign of the difference to determine monotonicity.

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 $n \ge 1$.