# Proving AM-GM with induction

I am trying to prove AM-GM with the following steps:

1. Prove that AM-GM holds for two variables
2. Prove that if AM-GM holds for $k$ variables then it holds for $2k$ variables
3. Prove that if AM-GM holds for $k$ variables then it holds for $k-1$ variables

If I prove all of these, I will have proved AM-GM. I have proved (1) and (2), but I am still struggling with (3).

Proof for (1)

By the Trivial Inequality, we know that the first statement is true. Therefore \begin{align} (a-b)^2 \ge 0 \implies\\ a^2-2ab+b^2 \ge 0 \implies \\ a^2+2ab+b^2 \ge 4ab \implies\\ a+b \ge 2\sqrt{ab} \implies\\ \frac{a+b}{2} \ge \sqrt{ab}\hspace{3mm}\blacksquare\\ \end{align}

Proof for (2)

Let our numbers be $a_1, a_2, a_3, \ldots, a_{2k}$. Denote $$x = \frac{\displaystyle\sum_{i=1}^{k}a_i}{k}, y = \frac{\displaystyle\sum_{i=k+1}^{2k}a_i}{k}$$ We know that by AM-GM for 2 variables, that $$\frac{x+y}{2}\ge \sqrt{xy}$$ By our definition of $x, y$ this means that $$\frac{x+y}{2} = \frac{\frac{\displaystyle\sum_{i=1}^{2k}a_i}{k}}{2}=\frac{a_1+a_2+a_3+\ldots+a_{2k}}{2k}$$ Again by definition, $$\sqrt{xy} = \sqrt{\left(\frac{x_1+x_2+\ldots+x_k}{k}\right)\left(\frac{x_{k+1}+x_{k+2}+\ldots+x_{2k}}{k}\right)}$$ By AM-GM for $k$ variables, $$\sqrt{\left(\frac{x_1+x_2+\ldots+x_k}{k}\right)\left(\frac{x_{k+1}+x_{k+2}+\ldots+x_{2k}}{k}\right)} \ge \sqrt{\sqrt[k]{x_1x_2\ldots x_k}\sqrt[k]{x_{k+1}x_{k+2}\ldots x_{2k}}}$$ This is simply equal to $$\sqrt[2k]{x_1x_2x_3\ldots x_{2k}}\hspace{3mm}\blacksquare$$

However, I have no idea how to proceed on (3). I tried using the definition of AM and GM but did not come up with anything useful. Feel free to point out any errors or flaws on my proofs for (1) and (2).

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 $$\sqrt[k]{x_1...x_{k-1}\sqrt[k-1]{x_1...x_{k-1}}}=\sqrt[k-1]{x_1...x_{k-1}}$$