Prove that the general arithmetic-geometric mean inequality \begin{equation*} (a_{1}a_{2}...a_{n})^\frac{1}{n}\leq\frac{a_{1}+a_{2}+...+a_{n}}{n} \end{equation*} holds for all $a_{i}$ positive real numbers.
I keep getting stuck half way. This is review material for me (which I feel like I should be getting easily, but that's not the case unfortunately).
Wikipedia has the solution to this problem:
If you would like to attempt it again before looking at the solution, here is a hint:
Attempt a proof by induction (as usual, consisting of the base case, hypothesis, induction step and then a conclusion). The base case is $n=1$. For the hypothesis, choose some non-negative real number $n$. For the induction step, rearrange the inequality and write \begin{equation*} \frac{a_1+...+a_n+a_{n-1}}{n+1}-(a_1...a_na_{n-1})^{\frac{1}{n+1}}\geq 0. \end{equation*} If you consider the quantity on the left as a function $f$, then the problem reduces to analysing the critical points of $f$ using tools from calculus. Is this okay? You said you got stuck half-way so I am happy to go through anything in more detail if you like.
Here is one approach which may or may not work:
Partial differential of the left hand side wrt $a_k$:
$$\frac{(a_k)^{-(n-1)/n}}{n} \prod_{i \neq k}(a_i)^{1/n}$$
and the right hand side:
$\frac{1}{n}$, now one could try to show this is bigger than the other.
http://math.stackexchange.com/questions/552780/is-tran-geq-nn-deta-for-a-symmetric-positive-definite-matrix-a-in-m/1095152#1095152
– Jesus RS Jan 07 '15 at 12:22