How to prove $(xyz)^{1/3} \le (x+y+z)/3$ using linear algebra

Question

To elaborate on the title, you can prove that $$\sqrt{xy}\le \frac{x+y}{2}$$ in the following way:

Is there a way that this can be extended to the inequality in the title and to a general case? For example for cube root? I tried the same thing done in the picture with three vectors but only got $$(xyz)^{{1\over 3}}\le \frac{x+y+z}{3^{1\over 3}}$$ which is obviously wrong.

Answer 1

I don't know exactly what to do to obtain a geometric proof (of course many proofs exist e.g. using convexity of the exponential function), but to relate the quantities via linear algebra I suggest $$ \det\pmatrix{x & y & z\\ y& z& x\\ z & x & y} =3xyz-(x^3+y^3+z^3).$$

Answer 2

Let $\text{AM-GM}_n$ stands for the statement of the arithmetic-geometric inequality in $n$ variables. A classical, elementary approach (due to Cauchy, Cours d'Analyse) is to show

$$ \text{AM-GM}_2 \vdash \text{AM-GM}_4 \vdash\text{AM-GM}_3.$$ Should I provide further details, or you prefer to figure them out by yourself?

Answer 3

By Cauchy-Schwarz inequality with

• $u=\left(\sqrt[3] {xy^\frac12},\sqrt[3] { yz^\frac12}, \sqrt[3] {zx^\frac12}\right)$

• $v=\left(\sqrt[3] {y^\frac12z},\sqrt[3] {xz^\frac12}, \sqrt[3] {x^\frac12y}\right)$

we obtain

$$uv\le |u||v| \iff 3\sqrt[3]{xyz}\le\sqrt{\left(\sqrt[3] {x^2y}+\sqrt[3] {y^2z}+\sqrt[3] {z^2x}\right)\left( \sqrt[3]{yz^2}+\sqrt[3]{x^2z}+\sqrt[3]{xy^2}\right)}\le \sqrt{\left(\sqrt[3] {x^3}+\sqrt[3] {y^3}+\sqrt[3] {z^3}\right)\left( \sqrt[3]{z^3}+\sqrt[3]{x^3}+\sqrt[3]{y^3}\right)}=x+y+z$$

Answer 4

An Identity which Proves the Inequality $$ \begin{align} &\frac12\left((x-y)^2+(y-z)^2+(z-x)^2\right)(x+y+z)\\ &+3x(y-z)^2+3y(z-x)^2+3z(x-y)^2\\[4pt] &=(x+y+z)^3-27xyz \end{align} $$ and the inequality follows.

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Motivation Behind the Identity

Knowing that the extreme points of $(x+y+z)^3-27xyz$ are when $x=y=z$, we should look at $$ \begin{align} &\frac12\left((x-y)^2+(y-z)^2+(z-x)^2\right)\\ &=x^2+y^2+z^2-xy-yz-zx\\[4pt] &=(x+y+z)^2-3(xy+yz+zx) \end{align} $$ Multiply by $x+y+z$ $$ \begin{align} &\frac12\left((x-y)^2+(y-z)^2+(z-x)^2\right)(x+y+z)\\ &=(x+y+z)^3-3(xy+yz+zx)(x+y+z)\\[4pt] &=(x+y+z)^3-9xyz-3xy^2-3xz^2-3yz^2-3yx^2-3zx^2-2zy^2\\[4pt] &=(x+y+z)^3-27xyz-3x(y-z)^2-3y(z-x)^2-3z(x-y)^2 \end{align} $$ and the identity follows.