<s>[INST] Prove $((a+b)/2)^n\leq (a^n+b^n)/2$ 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.
 [/INST] $$\left(\frac{a+b}{2}\right)^n = \frac{ \sum_{k=0}^n \binom{n}{k} a^{n-k}b^k}{2^n}= \frac{ \sum_{k=0}^n \binom{n}{k} a^{k}b^{n-k}}{2^n}$$
Therefore
$$\left(\frac{a+b}{2}\right)^n =\frac{1}{2} \frac{ \sum_{k=0}^n \binom{n}{k} (a^{n-k}b^k+a^kb^{n-k})}{2^n}$$
Now for each $k$ we have by AM-GM:
$$a^{n-k}b^k \leq\frac{a^{n}+...+a^{n}+b^n+..+b^n}{n}=\frac{(n-k)a^{n}+kb^n}{n} $$
and similarly
$$a^{k}b^{n-k} \leq\frac{a^{n}+...+a^{n}+b^n+..+b^n}{n}=\frac{ka^{n}+(n-k)b^n}{n} $$
Therefore, by adding them together we get
$$a^{n-k}b^k+a^kb^{n-k}\leq a^n+b^n$$
This yields
Therefore
$$\left(\frac{a+b}{2}\right)^n =\frac{1}{2} \frac{ \sum_{k=0}^n \binom{n}{k} (a^{n-k}b^k+a^kb^{n-k})}{2^n} \leq\frac{1}{2} \frac{ \sum_{k=0}^n \binom{n}{k} (a^{n}+b^{n})}{2^n}\\=\frac{a^n+b^n}{2} \frac{ \sum_{k=0}^n \binom{n}{k} }{2^n} =\frac{a^n+b^n}{2} $$
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