Answer
See explanations.
Work Step by Step
Step 1. Let $m=n-3$, as $n\geq3$, we have $m\geq0$ and the original statement becomes $(m+4)^2\lt2(m+3)^2$. We will prove this new statement which is equivalent to the original statement.
Step 2. Prove the statement is true when $m=0$: $(4)^2\lt2\times3^2$ or $16\lt18$ which is true, thus it is true for $m=0$.
Step 3. Assume the statement is true when $m=k$ where $k\gt0$: we have $(k+4)^2\lt 2 (k+3)^2$ or $k^2+8k+16\lt 2k^2+12k+18$ which gives $k^2+4k+2\gt0$ (Note: this is obviously true as $k\gt0$)
Step 4. Prove it is true for $m=k+1$: We have (RHS stands for Right-Hand-Side)
$RHS-LHS=2(k+4)^2-(k+5)^2=2k^2+16k+32-(k^2+10k+25)=k^2+6k+11\gt0$ (the result is positive as $k\gt0$). Thus we have $LHS\lt RHS$ for $m=k+1$
Step 5. Conclusion: with mathematical induction, we have proved that the statement is true for all natural numbers $n\geq3$.
