Prove that $4^n-3^n\gt 2n^2$ for all $n\ge 3$

Question

I found this problem in a textbook, I confirmed it works when n = 3, and followed up with the inductive step, $$4^{n+1}-3^{n+1}\gt2(n+1)^2$$ but I'm stuck at $$4^n\cdot4-3^n\cdot3\gt2n^2+4n+2$$ Induction is a new thing for me, so please excuse any mistakes, thanks.

Answer 1

Suppose $2n^2<4^n-3^n$ and $n\geq 3$.

\begin{align} 2(n+1)^2&=2n^2+4n+2\\ &<8n^2\\ &<4(4^n-3^n)\\ &=4^{n+1}-4\cdot 3^n\\ &<4^{n+1}-3^{n+1}. \end{align}

Answer 2

Solution without induction.

By the Lagrange's theorem there is $\theta\in(0,1)$ for which also by Bernoulli we obtain: $$4^{n+1}-3^{n+1}=\frac{4^{n+1}-3^{n+1}}{4-3}=(n+1)(3+\theta)^{n+1}>$$ $$>(n+1)3^{n+1}>(n+1)(1+2(n+1))>2(n+1)^2.$$