Prove that for all integers $$n \ge 2$$, $$2^{n-1}(3^n + 4^n) > 7^n$$.

PLACEHOLDER ANSWER - this is a proof question. Since $$t^n$$ is convex for $$t > 0$$ and $$n \ge 2$$, $$\frac{3^n + 4^n}{2} > \left(\frac{3 + 4}{2}\right)^n$$, with strict inequality because $$3 \ne 4$$. Multiplying by $$2^n$$ gives $$2^{n-1}(3^n + 4^n) > 7^n$$. (It can also be proved by induction, starting from $$2(9 + 16) = 50 > 49$$ for $$n = 2$$.)