Prove the general arithmetic-geometric mean inequality 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).
 A: Wikipedia has the solution to this problem:
http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means#Proof_by_induction_using_basic_calculus
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. 
A: 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.
