Struggling with this proof.
Prove that $$\left(\frac{a+b}{2}\right)^n≤\frac{a^n+b^n}{2},$$ where $a$ and $b$ are real numbers such that $a+b≥0$ and $n$ is a positive integer.
What technique would you use to prove this (e.g. induction, direct, counter example). How would you go about proving it?
Thanks in advance.
Best Answer
Solution 1.(Partial solution) $x^n$ is a convex function on $\mathbb{R}^+$ for $n\geq 1$, thus by Jensen: $$\left(\frac{a+b}{2}\right)^n\leq \frac{a^n+b^n}{2}$$
Solution 2.('Hidden' usage of condition) For $n=1$, true. By induction: $$\left(\frac{a+b}{2}\right)^n=\left(\frac{a+b}{2}\right)^{n-1}\left(\frac{a+b}{2}\right)\leq \frac{a^{n-1}+b^{n-1}}{2}\cdot\frac{a+b}{2} $$ And: $$\frac{a^n+b^n}{2}-\frac{a^{n-1}+b^{n-1}}{2}\cdot\frac{a+b}{2}=\frac{(a^{n-1}-b^{n-1})(a-b)}{4}$$ And the factors have the same sign.
Solution 3. Let $\alpha=\frac{a+b}{2}$, then $a=\alpha+x$ and $b=\alpha-x$, then the LHS is $\alpha^n$, and the RHS-LHS is: $$ \frac{1}{2}(a^n+b^n)-\left(\frac{a+b}{2}\right)^n=\sum_{i=1}^{\lfloor n/2\rfloor} \binom{n}{2i}\alpha^{n-2i}x^{2i} $$ And by the conditions, all the terms are positive.