1. **Problem:** Prove that $n^2 - 7n + 12 \geq 0$ for all positive integers $n \geq 2$ using mathematical induction. 2. **Step 1: Base case** ($n=2$): Calculate $2^2 - 7 \times 2 + 12 = 4 - 14 + 12 = 2 \geq 0$. The base case holds. 3. **Step 2: Inductive hypothesis** Assume the statement is true for some integer $k \geq 2$, that is, $k^2 - 7k + 12 \geq 0$. 4. **Step 3: Inductive step** Show the statement holds for $k+1$: $$ (k+1)^2 - 7(k+1) + 12 = k^2 + 2k + 1 - 7k - 7 + 12 = k^2 - 5k + 6 $$ Rewrite: $$ k^2 - 5k + 6 = (k^2 - 7k + 12) + 2k - 6 $$ By the inductive hypothesis, $k^2 - 7k + 12 \geq 0$, so: $$ k^2 - 5k + 6 \geq 2k - 6 $$ For $k \geq 2$: $$ 2k - 6 \geq 2 \times 2 - 6 = 4 - 6 = -2 $$ But since we want to prove non-negativity, check the original expression at $k=2$: $$ (k+1)^2 - 7(k+1) + 12 $$ at $k=2$ is $3^2 - 7 \times 3 + 12 = 9 - 21 + 12 = 0 \geq 0$. Test the expression from $k=2$ onwards — factoring the quadratic: $$ n^2 - 7n + 12 = (n-3)(n-4) $$ The product is zero at $n=3$ and $n=4$, positive for $n \geq 4$ or $n \leq 3$ outside of $3 < n < 4$ which does not contain positive integers. Since we start at $n=2$, and $$ (2-3)(2-4) = (-1)(-2) = 2 \geq 0 $$ Therefore, the quadratic expression is non-negative for all integers $n \geq 2$. Hence by induction, the statement is true. **Final answer:** $$ n^2 - 7n + 12 \geq 0 \text{ for all } n \geq 2. $$