Proving AM-GM with induction
Solution 1:
To go from $k$ to $k-1$ just pick the last variable as the geometric mean of the other $k-1$.
Suppose that $$ \frac{x_1+...+x_k}{k} \geq \sqrt[k]{x_1...x_k}$$ Now choose $x_k = \sqrt[k-1]{x_1...x_{k-1}}$.
Then you'll get exactly what you want once you note that $$ \sqrt[k]{x_1...x_{k-1}\sqrt[k-1]{x_1...x_{k-1}}}=\sqrt[k-1]{x_1...x_{k-1}}$$