Q.Prove that $3^n \ge 1 + 2n$ for every natural number $n$.

[!TLDR] Triple the inductive hypothesis and note $4k\ge0$. [!ANSWER] Proved: $3^n\g …

Let $P(n)$: $3^n\ge1+2n$. Base case: $n=1$: $3^1=3$ and $1+2(1)=3$, so $3\ge3$ and $P(1)$ holds. Inductive step: assume $P(k)$: $3^k\ge1+2k$ for some $k\ge1$. Multiplying both sides by 3 (positive, so the inequality direction is preserved): $3^{k+1}=3\cdot3^k\ge3(1+2k)=3+6k$. We need this to be at least $1+2(k+1)=2k+3$. Since $3+6k\ge2k+3 \iff 4k\ge0$, which is true for every …

Inequality-by-induction: multiply the inductive hypothesis by 3, then verify $3+6k\ge2k+ …

Multiplying the inductive hypothesis by 3 but then comparing to the wrong target -- the target for $P(k+1)$ is $1 …