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The proof uses mathematical induction. First, the base case is verified by checking that $a_1=1 \ge \frac{1}{2}$.

Conditions: The sequence is defined by $a_{n+1}=\frac{1}{3}(a_n+1)$ for $n \ge 1$.; The initial value is $a_1=1$.; The goal is to prove $a_n \ge \frac{1}{2}$ for all $n \ge 1$.

The proof uses mathematical induction. First, the base case is verified by checking that $a_1=1 \ge \frac{1}{2}$.

Conditions: The sequence is defined by $a_{n+1}=\frac{1}{3}(a_n+1)$ for $n \ge 1$.; The initial value is $a_1=1$.; The goal is to prove $a_n \ge \frac{1}{2}$ for all $n \ge 1$.

Assume $P(k)$ is true: $2 + 4$ + ... + $2k = k(k+1)$. To prove $P(k+1)$, write the LHS as the sum up to 2k plus the next term $2(k+1)$.

Conditions: Inductive hypothesis: $P(k)$ is true.; Goal: Prove $P(k+1)$ is true.; Algebraic manipulation involves factoring common terms.

Assume $P(k)$ is true: $2 + 4$ + ... + $2k = k(k+1)$. To prove $P(k+1)$, write the LHS as the sum up to 2k plus the next term $2(k+1)$.

Conditions: Inductive hypothesis: $P(k)$ is true.; Goal: Prove $P(k+1)$ is true.; Algebraic manipulation involves factoring common terms.