Prove that for all $n\in \mathbb{N}$, it holds that for all real numbers $a,b\geq 0$, that $(\frac{a+b}{2})^n \leq \frac{a^n+b^n}{2}$

Question

I'm pretty new to proof by induction and I just wanted to see how I would prove something with multiple variables. Thanks.

Prove that for all $n\in \mathbb{N}$, it holds that for all real numbers $a,b\geq 0$, that $(\frac{a+b}{2})^n \leq \frac{a^n+b^n}{2}$

Answer 1

Since $a\mapsto ca,\,b\mapsto cb$ multiplies both sides by $c^n$, assume wlog that $a+b=2$ (unless $a+b=0$, but that case is trivial). Then we just need to prove $a^n+(2-a)^n\ge2$. For the inductive step, note$$\begin{align}a^{k+1}+(2-a)^{k+1}-a^k-(2-a)^k&=a^k(a-1)+(2-a)^k(1-a)\\&=(a^k-(2-a)^k)(a-1)\\&\ge0\end{align}$$completes the inductive step.

Answer 2

Base Case: $n = 1$. We have $\left(\frac{a+b}{2}\right)^1 = \frac{a^1 + b^1}{2}$.

Inductive step: Assume for some $k$ the conclusion holds, that is, there exists $k$ so that $\left(\frac{a+b}{2}\right)^k \leq \frac{a^k + b^k}{2}$. We show it holds for $k+1$. Note that \begin{align*} \left(\frac{a+b}{2}\right)^{k+1} = \left(\frac{a+b}{2}\right)^k\left(\frac{a+b}{2}\right) \leq \frac{a^k+b^k}{2}\left(\frac{a+b}{2}\right) = \frac{a^{k+1}+a^kb+ab^k + b^{k+1}}{4} = \frac{a^{k+1} + b^{k+1}}{4} + \frac{ab^k + ba^k}{4} \leq \frac{a^{k+1} + b^{k+1}}{4} + \frac{a^{k+1} + b^{k+1}}{4} = \frac{a^{k+1} + b^{k+1}}{2}, \end{align*} as desired. Where the last inequality follows by AM-GM.

Answer 3

Hint: $\left(\dfrac{a+b}{2}\right)^{n+1} = \left(\dfrac{a+b}{2}\right)^{n}\left(\dfrac{a+b}{2}\right) \le \left(\dfrac{a^n+b^n}{2}\right)\left(\dfrac{a+b}{2}\right)$