Answer
See explanations.
Work Step by Step
Step 1. Let $m=n-100$, as $n\geq100$, we have $m\geq0$. The original statement becomes $100(m+100)\leq (m+100)^2$. We will prove this new statement which is equivalent to the original.
Step 2. Prove the statement is true when $m=0$: $100(100)=(100)^2$ , thus it is true for $m=0$.
Step 2. Assume the statement is true when $m=k$: we have $100(k+100)\leq (k+100)^2$
Step 3. Prove it is true for $m=k+1$: we have $LHS-RHS=100(k+101)- (k+101)^2=100k+10100-k^2-202k-10201=-k^2-102k-101$. As $k\geq0$, we have $LHS-RHS\leq0$ and $LHS\leq RHS$.
Thus, the statement is also true for $m=k+1$
Step 4. Conclusion: with mathematical induction, we have proved that the statement is true for all natural numbers $n\geq100$.
