# Proving the AM-GM Inequality with a given fact

Given $$x + y + z \geq 3$$ for all $$(x, y, z) \in \mathbb{R}^{3}$$ such that $$x,y, z > 0$$ provided $$xyz = 1$$, show that

$$\frac{a_1+a_2+a_3}{3} \geq \sqrt[3]{a_1a_2a_3}$$ holds.

I'm not really sure how to do this; the only relation I see to use is $$x + y + z \geq 3$$ in the denominator of the AM term. But even if we did prove this, wouldn't it only be proving it for the case where $$a_1a_2a_3 = 1$$?

Note that the first part of the exercise was to prove that $$x + y + z \geq 3$$ holds for $$x,y,z > 0$$ provided $$xyz=1$$. The way to do this was with lagrange multipliers. I've already done this, but maybe the second part uses lagrange multipliers as well.

• Hint: Apply your fact to $(x,y,z) = \left( \frac{a_1}{g}, \frac{a_2}{g},\frac{a_3}{g}\right)$ where $g = \sqrt[3]{a_1a_2a_3}$. – achille hui Apr 19 at 13:16

## 1 Answer

We must assume that $$a_1, a_2, a_3$$ are non-negative, otherwise the AM-GM inequality does not necessarily hold (e.g. for $$a_1 = 2, a_2 = a_3 = -1$$).

Now you can prove the AM-GM inequality using the “given fact” as follows:

If any $$a_j$$ is zero then the inequality holds trivially. Otherwise set $$x = \frac{a_1}{\sqrt[3]{a_1a_2a_3}} \, , \, y = \frac{a_2}{\sqrt[3]{a_1a_2a_3}} \, , \, z = \frac{a_3}{\sqrt[3]{a_1a_2a_3}} \, .$$ Then $$xyz = 1$$ and therefore $$3 \le x + y + z = \frac{a_1 + a_2 + a_3}{\sqrt[3]{a_1a_2a_3}}$$