# GM-AM Inequality Proof

I encountered this problem in Kuratowski's intro to calculus. I'd like help in the direction of the hint provided in the book, please. In the hint, I don't understand how the geometric mean is greater than $G$, or where it's even headed. Thank you.

Denote by $A$ and $G$ the arithmetic and geometric means of the numbers $a_{1},a_{2},...,a_{n},$ respectively, i.e., $$A=\frac{a_{1}+a_{2}+...a_{n}}{n},\ \ \ \ \ \ \ \ \ G=\sqrt[\leftroot{-2}\uproot{2}n]{a_{1}\cdot a_{2}\cdot ...\cdot a_{n}}.$$ Prove that if the numbers $a_{1},a_{2},...,a_{n}$ are positive, then $G\le A$.

Hint: Precede the proof by the remark that if $a_{1}<A<a_{2}$ then the arithmetic mean of the numbers $A,a_{1}+a_{2}-A,a_{3},...,a_{n}$ is $A$ and the geometric mean of these numbers is $>G$.

By following the hint I get that $$A(a_{1}+a_{2}-A)>a_{1}a_{2}$$ How is this true?

Here's where the hint is headed. If the numbers in your list are not all equal, then there exist two numbers, call them $a_1$ and $a_2$, such that $A$ lies strictly between them. Create a new list from the original list by replacing $a_1$ with $A$, and replacing $a_2$ with $a_1+a_2-A$. Then, according the the hint (which you've proved via @dxiv's hint), the new list has the same arithmetic mean as, but larger geometric mean than, the original. What next? You can repeat this procedure until you obtain a list consisting of all $A$. What can you say about the arithmetic mean and geometric mean of this final list, and how does these relate to the original list?
By following the hint I get that: $\quad A(a_{1}+a_{2}-A)>a_{1}a_{2}$
Next hint:   that's equivalent to $\,(A-a_1)(A-a_2) \lt 0\,$.