Proving AM ≥ GM in 3 variables using the Cauchy-Schwarz inequality This question is from the textbook Introduction to Linear Algebra by Gilbert Strang, where the author has asked to prove that
$$\sqrt[3]{xyz} \le \frac{x+y+z}{3}. $$
The only equations at hand are the Schwarz inequality:
$$\newcommand{\norm}[1]{\left\lVert#1\right\rVert}
\left | \vec u \ {\large\cdot} \ \vec v \right| \le \norm{\vec u} \ \norm{\vec v},$$
and the Triangle inequality:
$$ \norm{\vec u +\vec v} \le \norm{\vec u} + \norm{\vec v}.$$
For only two variables $x$ and $y$, I can prove it in the following way:
Let's say $\vec u = (x,y)$ and $\vec v = (y,x)$.
Then by the Schwarz inequality,
$$\left | 2xy \right | \le x^2 + y^2 \implies \sqrt{x^2y^2} \le \frac{x^2+y^2}{2}$$
But I am unable to extend it to 3 variables. Could you please help.
 A: Cauchy Schwarz implies $xy + yz + zx \le x^2 + y^2 + z^2$ and for positive arguments $0 \le x + y + z$,
so that $0 \le (x+y+z)(x^2 + y^2 + z^2 - xy -yz -zx)$ implies
$3xyz \le x^3 + y^3 + z^3$, substitute $x$,$y$ and $z$ with $\sqrt[3]{x}$,$\sqrt[3]{y}$ and $\sqrt[3]{z}$ respectively to get $\sqrt[3]{xyz} \le \frac{x+y+z}{3}$
A: You can use the Cauchy-Schwarz inequality to prove that
$$\frac{x + y}{2} \geqslant \sqrt{xy} \tag{$\dagger$}$$
as you have done, if you replace $x^2$ and $y^2$ with $x$ and $y$.
We shall extend the AM–GM inequality from two to four variables, then reduce it to three. Applying $(\dagger)$ twice on the arithmetic mean of four variables, $(w + x + y + z)/4$, gives
$$\frac{(w + x) + (y + z)}{4} \geqslant \frac{\sqrt{wx} + \sqrt{yz}}{2}\geqslant \sqrt[4]{wxyz}.$$
Now let $w = \frac{1}{3}(x + y + z)$ so that
$$\begin{align*}
(x + y + z)/3 &\geqslant \sqrt[4]{(x + y + z)/3 \cdot xyz}\\
\frac{1}{3^{3/4}}(x + y + z)^{3/4} &\geqslant \sqrt[4]{xyz}\\
\frac{x + y + z}{3} &\geqslant \sqrt[3]{xyz},\\
\end{align*}$$
as desired.
