prove AM-GM inequality for $n=k-1$

Question

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.

Answer 1

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.