Prove that $\frac{(a+1)^k}{a^k+1} \leq 2^{k-1}$

Question

if $ k \geq 0$ Prove that $\frac{(a+1)^k}{a^k+1} \leq 2^{k-1}$

Proof (induction)

let k = 1 then $$\frac{(a+1)^1}{a^1+1} \leq 2^{1-1}$$ , implies that $$ 1 \leq 1$$ so p(1) is true.

Inductive step

$p(l)$ : $\frac{(a+1)^l}{a^l+1} \leq 2^{l-1}$

$p(l+1)$ : $\frac{(a+1)^k{l+1}}{a^{l+1}+1}$ this is where i am having a dificulty to show that expression is greater or equal to $2^l$. any idea will be greatly appreciate it

Answer 1

The inequality of arithmetic and k-th degree mean states that $$ \frac{a_1+a_2+…+a_n}n\le \sqrt[k]{\frac{a_1^k+a_2^k+…+a_n^k}{n}} $$ or $$ \frac{(a_1+a_2+…+a_n)^k}{a_1^k+a_2^k+…+a_n^k}\le n^{k-1} $$ Now set $n=2$ etc.

Answer 2

$$\frac{(a+1)^k}{a^k+1} \leq 2^{k-1}\iff\left(\frac{a+1}{2}\right)^k\leq\frac{a^k+1}{2}$$

If $k\in\mathbb{N}$, $f(x)=x^k$ is convex, i.e. $f(\lambda x + (1-\lambda)y)\le\lambda f(x)+(1-\lambda)f(y)$ for all $x,y$, and $\lambda\in[0,1]$.

Set $x=1,y=a,\lambda=\frac{1}{2}$ to obtain the desired result.