AM/GM mean for $n=3$ using $n=4$

Question

Suppose I know that $\frac{a+b+c+d}{4}\ge(abcd)^\frac{1}{4}$ for $a$, $b$, $c$ and $d \in \mathbb{R^+}$

How can I show that $\frac{x+y+z}{3}\ge(xyz)^\frac{1}{3}$ for $x$, $y$, $z \in \mathbb{R^+}$

My first guess was to let $d=1$ and go from there but that didn't work out. Any tips?

For the general case, see for example https://math.stackexchange.com/questions/739804/how-to-understand-cauchys-proof-of-am-gm-inequalitythe-last-step. — Martin R, Feb 11 '18 at 17:18

score 1 · Accepted Answer · answered Feb 11 '18 at 16:35

You set $d=\sqrt[3]{abc}$. Then, an easy calculation shows $\sqrt[4]{abcd}=\sqrt[4]{d^3\cdot d}=d=\sqrt[3]{abc}$, and from $$\frac{a+b+c+\sqrt[3]{abc}}4\ge\sqrt[3]{abc}$$ follows immediately $$\frac{a+b+c}3\ge\sqrt[3]{abc}.$$

AM/GM mean for $n=3$ using $n=4$

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