<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] A much easier (than my original) answer
Let $c>0$ and let $$p_c(x)=(c+x)^n +(c-x)^n=2\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}c^{n-2k}x^{2k}$$ Note that $p_c(x)$ has only positive coefficients and every term has even power. So $p_c(0)\leq p_c(x)$ for all $x$.
Then let $c=\frac{a+b}{2}$ and $x=\frac{a-b}{2}$. Then $c+x=a$ and $c-x=b$, and we get:
$$a^n + b^n =p_c(x)\geq p_c(0) = c^n + c^n=2\left(\frac{a+b}{2}\right)^n$$

My original answer
Preliminary
Let $$f(x,y)=\sum_{i=0}^{n-1} x^{i}y^{n-1-i}.$$ Then we can easily show that $$f(x,y)=f(y,x)\tag{1}$$
$$(x-y)f(x,y)=x^n-y^n\tag{2}$$
$$f(x,y_1)\leq f(x,y_2)\text{ when } x\geq 0\text{ and } |y_1|\leq y_2\tag{3}$$
Proof
Let $a\leq b$ and let $c=\frac{a+b}{2}$. Note $b-c=c-a\geq0.$
Now, since $a\leq b$ and $-a< b$, we have $|a|\leq b$ so we have, by $(1)$ and $(3)$:
$$f(c,a)\leq f(c,b)=f(b,c)$$
Multiplying this inequality by $c-a=b-c\geq 0$, we get, by $(2)$:
$$c^n-a^n\leq b^n-c^n.$$
Hence $$\frac{a^n+b^n}{2} \geq c^n,$$
which is the inequality we want.
Sadly, this doesn't work for non-integers $n\geq 1$.
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