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.

  • 1
    $\begingroup$ Hint: Apply your fact to $(x,y,z) = \left( \frac{a_1}{g}, \frac{a_2}{g},\frac{a_3}{g}\right)$ where $g = \sqrt[3]{a_1a_2a_3}$. $\endgroup$ – achille hui Apr 19 at 13:16

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 $$ 3 \le x + y + z = \frac{a_1 + a_2 + a_3}{\sqrt[3]{a_1a_2a_3}} $$


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