TITLE: prove AM-GM inequality for $n=k-1$
QUESTION [1 upvotes]: The question is assume the AM-GM inequality holds for $n$, prove for $n=k-1$.
After a bit of moving, I'm stuck at the following.
$\frac{x_1+...+x_{k-1}}{k-1}\geq \frac{k}{k-1}x_k^{1/k} ((x_1...x_{k-1})^{1/k}-x_k^\frac{k-1}{k})$
What do I do next? Also please elaborate if you are giving hint.

REPLY [1 votes]: Hint: let
\begin{eqnarray*}
X=x_k= \frac{x_1+\cdots+x_{k-1}}{k-1}.
\end{eqnarray*}
We have
\begin{eqnarray*}
 \frac{x_1+\cdots+x_{k}}{k}= \frac{x_1+\cdots+x_{k-1}+ \frac{x_1+\cdots+x_{k-1}}{k-1}}{k} =\frac{x_1+\cdots+x_{k-1}}{k-1} =X.
\end{eqnarray*}
So
\begin{eqnarray*}
X \geq \sqrt[k] {x_1 \cdots x_{k-1} X}
\end{eqnarray*}
raise this to the power of $k$, divivde through by $X$ & then take the $(k-1)^{th}$ root.
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