show this inequality $3^{\frac{n}{4}}\cdot (3^{\frac{1}{4}}-1)^{8-n}\le 1+n$

Question

let $n\in Z$,and $n\in[0,8]$,show that $$3^{\frac{n}{4}}\cdot (3^{\frac{1}{4}}-1)^{8-n}\le 1+n$$

it seem use Jenson inequality,But How to use it?Thanks.

Answer 1

Just check this inequality for $n\in[0,8]$.

Also by the dxiv's beautiful hint we can make the following thing.

Let $f(x)=3^{\frac{x}{4}}.$

Hence, $f$ is a convex function.

But a graph of $f$ and a line $y=1+x$ have two common points: $(0,1)$ and $(8,9)$.

Thus, $3^{\frac{x}{4}}\leq x+1$ by definition of convexity, which says

$$3^{\frac{n}{4}} (3^{\frac{1}{4}}-1)^{8-n}\le 1+n$$ because $0<3^{\frac{1}{4}}-1<1$.

Answer 2

Hint: $\;3^{n/4} \le 1 + n\,$ for $\,0 \le n \le 8\;$ (and $n$ doesn't even need to be an integer).