I'm about to prove that for any $a,b>0$ and $n\in\mathbb{N},$ the inequality: $\frac{a^n+b^n}{2}\geq\left(\frac{a+b}{2}\right)^n$ holds.
By induction I get: $$\left(\frac{a+b}{2}\right)\cdot\frac{a^n+b^n}{2}$$ $$=\frac{1}{2}\cdot\frac{a^{n+1}+b^{n+1}+ab^n+ba^n}{2}$$ $$=\frac{1}{2}\left(\frac{a^{n+1}+b^{n+1}}{2}+\frac{ab^n+ba^n}{2}\right).$$ Now I have to prove that $$\frac{1}{2}\left(\frac{a^{n+1}+b^{n+1}}{2}+\frac{ab^n+ba^n}{2}\right)$$ $$\leq\frac{a^{n+1}+b^{n+1}}{2}$$ $$\Leftrightarrow a^{n+1}+b^{n+1}\geq ab^n +ba^n.$$ But I don't know how to prove that one, is it possible to be proved by induction ? Thank you.
Answer
Your desired inequality can be written as $0\leq (a-b)(a^n-b^n)$ this is true regardless which is bigger.
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