We will prove by induction that $2^n + 3^n < 6^n$ for all $n \geq 1$.

Step 1: Base Case

For $n = 1$: $2^1 + 3^1 = 2 + 3 = 5$ $6^1 = 6$ Clearly, $5 < 6$, so the inequality holds for $n = 1$.

Step 2: Induction Hypothesis

Assume that the inequality $2^n + 3^n < 6^n$ holds for some $n = k$. That is: $2^k + 3^k < 6^k$ We will now prove that the inequality also holds for $n = k + 1$.

Step 3: Inductive Step

We need to show that: $2^{k+1} + 3^{k+1} < 6^{k+1}$

From the properties of exponents, we know: $6^{k+1} = 6 \cdot 6^k$ Now, express $2^{k+1}$ and $3^{k+1}$ using the same properties: $2^{k+1} = 2 \cdot 2^k \quad \text{and} \quad 3^{k+1} = 3 \cdot 3^k$ Thus, the inequality we need to prove becomes: $2 \cdot 2^k + 3 \cdot 3^k < 6 \cdot 6^k$

Using the induction hypothesis $2^k + 3^k < 6^k$, let's try bounding the left-hand side.

We know that: $2 \cdot 2^k + 3 \cdot 3^k < 2 \cdot 6^k + 3 \cdot 6^k = (2 + 3) \cdot 6^k = 5 \cdot 6^k$ Thus, we now need to show: $5 \cdot 6^k < 6 \cdot 6^k$ Clearly, this is true because $5 < 6$. Hence: $2^{k+1} + 3^{k+1} < 6^{k+1}$

Step 4: Conclusion

By mathematical induction, we have shown that $2^n + 3^n < 6^n$ for all $n \geq 1$.

Would you like more details on any part of this proof, or do you have any questions?

Here are five related questions:

1. Can you explain how induction works in more detail?
2. What is the significance of the base case in induction?
3. Can this inequality be extended for $n < 1$?
4. How would the proof change if the inequality were $2^n + 3^n \leq 6^n$?
5. Can you use strong induction to prove this inequality?

Tip: In proofs by induction, always clearly separate the base case, induction hypothesis, and inductive step to avoid confusion.