Why does proving $-\frac{2}{3}a_n+\frac{1}{3}\le0$ reduce to the already known bound $a_n\ge\frac{1}{2}$?

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 $n \ge 1$.

Reasoning, step by step

1. Start with the inequality to prove: $-\frac{2}{3}a_n+\frac{1}{3}\le0$.
2. Add $\frac{2}{3}a_n$ to both sides: $\frac{1}{3} \le \frac{2}{3}a_n$.
3. Multiply both sides by $\frac{3}{2}$ (or divide by $\frac{2}{3}$): $\frac{1}{2} \le a_n$.
4. Recognize that $a_n \ge \frac{1}{2}$ is the previously proven inductive claim.
5. Conclude that the difference is non-positive, implying the sequence is non-increasing.

Example

The board shows the boxed equivalence: $-\frac{2}{3}a_n+\frac{1}{3}\le0 \iff \frac{1}{3}\le\frac{2}{3}a_n \iff \frac{1}{2}\le a_n$.

Common misconceptions

• Forgetting that dividing or multiplying by a positive constant preserves the inequality direction.
• Assuming the equivalence only holds in one direction.

Watch the explanation

Explore next

• Does this example prove the sequence is non-decreasing or non-increasing?
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