I am reading a proof (see page 2) of the following statement. Let $n\in \mathbb{R}$ and suppose we are given real numbers $a_1\geq a_2\geq \dots a_n\geq 0$. Then
$$
\frac{1}{n}\sum_{i=1}^n a_i\geq \sqrt[n]{\prod_{i=1}^n a}.
$$
Until the part, where it says, right below the equation (1.1),
Now, according to our assumption, the AM-GM inequality holds for any collection of n
non-negative numbers. Hence, in particular, for the collection $a_2$, $a_3$, . . . , $a_n$,$a_1+a_{n+1}−A$.
The text marked with bold, especially the the term $a_1+a_{n+1}−A$ is what I want to ask about. The assumption can only be applied if $a_1+a_{n+1}−A$ lies somewhere between $a_2\geq \dots \geq a_n$, or outside these numbers (it is still non-negative). But, I can't check it myself whether or not it does. I mean, is $a_n\geq a_1+a_{n+1}−A$, or does it lie between two unknown consecutive terms $a_m,a_{m+1}$? I hope you understand what I am trying to ask. Can you help me to find the lower/upper bound of $a_1+a_{n+1}−A$?
Best Answer
Note that the author states
Because $a_1 - A$ and $a_{n+1}$ are non-negative, it follows that their sum $a_1 + a_{n+1} - A$ is non-negative.
We can verify that $a_1 \geq A$ as the author claims as follows. Note that $$ \begin{align} a_1 - A &= a_1 - \frac{a_1 + a_2 + \cdots + a_{n+1}}{n+1} \\ & = \frac{(n+1)a_1 - a_1 - a_2 - \cdots - a_{n+1}}{n+1} \\ & = \frac{(a_1 - a_2) + (a_1 - a_3) + \cdots + (a_1 - a_{n+1})}{n+1}. \end{align} $$ Because $a_1 \geq a_2 \geq \cdots \geq a_{n+1}$, it is clear that the numerator is positive, so that $a_1 - A \geq 0$ holds as desired.