<s>[INST] $3^n+4^n<5^n$ for all $n>2$ I'm doing the following induction proof and wanted to know if this was valid. I think it is, but I'm seeing more complicated solutions than what I did. What I did seems much easier.
Prove that $3^n+4^n<5^n$ for all $n>2$.
When $n=3$ we get $91<125$. No problem, now assume the result is true from $k<n$, $(5^k>3^k+4^k)$ and consider $5^{k+1}=5 \times 5^k>5(3^k+4^k)=5\times 3^k + 5\times 4^k>3\times 3^k+4\times 4^k=3^{k+1}+4^{k+1}$ since $5\times 3^k>3\times 3^k$ and $5\times 4^k> 4\times 4^k$.
 [/INST] Here's a method I like
for showing
$f(n) > g(n)$
for $n \ge n_0$.

*

*Show that
$f(n_0) > g(n_0)$.


*Show that,
if $n \ge n_0$
and $f(n) > g(n)$ then
$f(n+1)-f(n)
\ge g(n+1)-g(n)
$.
Then $f(n) > g(n)$
for $n \ge n-0$.
In this case,
$f(n) = 5^n,
g(n) = 3^n+4^n$.
Choose $n_0 = 3$.
$f(n_0) = 5^3=125,
g(n_0) = 3^3+4^3 = 91
\lt f(n_0)$.
So (1) holds.
If $n \ge 3$
and $f(n) > g(n)$ then
$f(n+1)-f(n)
=5^{n+1}-5^n
=4\cdot 5^n
\gt 4(3^n+4^n),\\
g(n+1)-g(n)
=2\cdot 3^n+3\cdot 4^n
\lt 3(3^n+4^n)
\lt f(n+1)-f(n)
$.
</s>