Prove that for all $n \in \mathbb{N}$: $$2^{n-1}(a^n + b^n) \ge (a + b)^n$$ I used induction (for $k = 1,2,...n-1$), and ended up with the following equation:
$$(a + b)(a + b)^{n-1} \le (a + b)(a^{n-1} + b^{n-1})2^{n-2}$$
Now, I have a feeling I'm on the right track, but I'm not totally sure. Could someone give me a pointer one where I should go from here? Any help would be greatly appreciated. Thanks!
You are almost there since the last equation you have can be written as $$(a + b)(a + b)^{n-1} \le (a + b)(a^{n-1} + b^{n-1})2^{n-2}=(a^n+b^n+ab^{n-1}+ba^{n-1})2^{n-2}\le(a^n+b^n+a^n+b^n)2^{n-2}=2(a^n+b^n)2^{n-2}=(a^n+b^b)2^{n-1}$$ where you need only to prove (so that the above holds) that $$ab^{n-1}+ba^{n-1}\le a^n+b^n$$ or equivalently that $$(a-b)(a^{n-1}-b^{n-1})\ge0$$ which can be (straightforwardly) done by selecting cases (whether $a>b$ or $a\le b$).
As $n\ge 1$, $x\to x^n$ is convex, so via Jensen inequality: $$ \left(\frac{a+b}2\right)^n \le \frac{a^n+b^n}2 $$ that gives the result.
You can write $a={r\over2}(1+x)$ and $b={r\over2}(1-x)$, which turns the inequality into
$${1\over2}\left((1+x)^n+(1-x)^n\right)\ge1$$
When the binomials are expanded, the odd powers of $x$ cancel.
