# 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).

• Try using the concavity of the logarithm. Other proofs can be found here : link Commented Sep 4, 2013 at 9:56
• Why is this tagged (calculus)? Do you want some solution using calculus in some way? Commented Sep 4, 2013 at 9:58
• I guess I should add "analysis" tag.. its a reviwe for my analysis course. and the prof said we should know these things, but i can't seem to get it.
– Dome
Commented Sep 4, 2013 at 10:01
• Have you looked at the proofs given on Wikipedia? Have you looked at other questions about this inequality? For example questions linked in this one. Commented Sep 4, 2013 at 10:06
• @MartinSleziak: Thank you!
– Dome
Commented Sep 4, 2013 at 10:17

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.

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.