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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.

Ben
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arindam mitra
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2 Answers2

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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.

Ben
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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 $\sqrt[3]{xyz} \le \frac{x+y+z}{3}$

r9m
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