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).

  • 2
    $\begingroup$ Try using the concavity of the logarithm. Other proofs can be found here : link $\endgroup$
    – user37238
    Commented Sep 4, 2013 at 9:56
  • $\begingroup$ Why is this tagged (calculus)? Do you want some solution using calculus in some way? $\endgroup$ Commented Sep 4, 2013 at 9:58
  • $\begingroup$ 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. $\endgroup$
    – Dome
    Commented Sep 4, 2013 at 10:01
  • $\begingroup$ 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. $\endgroup$ Commented Sep 4, 2013 at 10:06
  • $\begingroup$ @MartinSleziak: Thank you! $\endgroup$
    – Dome
    Commented Sep 4, 2013 at 10:17

2 Answers 2


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .