Elementary proof for $\sqrt[3]{xyz} \leq \dfrac{x+y+z}{3}$

Question

I am searching for an elementary proof of the AM-GM inequality in three variables:

$\sqrt[3]{xyz} \leq \dfrac{x+y+z}{3}$

The inequality of the geometric mean vs the arithmetic mean of two variables can be proven elementarily via

$x - 2 \sqrt{xy} + y \geq 0$

whenever $x,y > 0$. I am searching for a proof that uses a similar technique based on elementary arithmetic.

Answer 1

Let $A=\dfrac{x+y+z}{3}$ and $G=\sqrt[3]{xyz}$

Applying AM-GM inequality for the $4$ (you can easily prove it for $4=2^2$ quantities, in general for $2^n$ quantities) quantities , $x,y,z,A$; we have $$\sqrt[4]{xyzA} \leq \dfrac{x+y+z+A}{4}$$ $$\sqrt[4]{A} \cdot G^{\frac34} \leq \dfrac{3A+A}{4}$$ $$A^{\frac14} \cdot G^{\frac34} \leq A$$ $$G^{\frac34} \leq A^{\frac34}$$ $$G \leq A$$ $$\sqrt[3]{xyz} \leq \dfrac{x+y+z}{3}$$

Hope this helps.

Answer 2

For any $x,y,z > 0$, define $u = \sqrt[3]{x}, v = \sqrt[3]{y}, w = \sqrt[3]{z}$.
We have

$$\begin{align} \frac{x + y + z}{3} - \sqrt[3]{xyz} &= \frac13( u^3 + v^3 + w^3 - 3uvw)\\ &= \frac16 (u+v+w)((u-v)^2+(v-w)^2+(w-u)^2)\\ &\ge 0\end{align}$$

Furthermore, if $x,y,z$ are not all the same, at least one of $(u-v)^2$, $(v-w)^2$ or $(w-u)^2$ is positive. This means the inequality is strict unless $x = y = z$.

Answer 3

$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx) = \tfrac12(x+y+z)((x-y)^2+(y-z)^2+(z-x)^2) \ge 0$$

Answer 4

You can have an elementary proof of the general AM-GM inequality: $$|x_1\dotsm x_n|^{1/n}\le \frac{x_1+\dotsm +x_n}{n}\tag*{AG($n$)}$$ with the following non-standard form of induction:

• prove that $\,\mathrm{AG}(n)\implies \mathrm{AG}(2n)$,
• prove that $\,\mathrm{AG}(n)\implies \mathrm{AG}(n\color{red}{-}1)$.

Answer 5

An equivalent form is $$x^3+y^3+z^3\ge 3xyz\tag{*}$$ for $x$, $y$, $z\ge0$. Without loss assume that $0\le x\le y\le z$. Then $y=x+a$ and $z=x+a+b$ with $a$, $b$, $x\ge0$. Also $$x^3+y^3+z^3-3xyz=x^3+(x+a)^3+(x+a+b)^3-3x(x+a)(x+a+b) =3xa^2+3xab+3xb^2+2a^3+3a^2b+3ab^2+b^3\ge0.$$ Then (*) follows.

Answer 6

Since our inequality is homogeneous, we can assume that $xyz=1$ and we need to prove that $$x+y+z\geq3.$$

Now, since $xyz=1$ there are two numbers $a$ and $b$ from $\{x,y,z\}$ for which $a\geq1$ and $b\leq1$.

Indeed, if $x>1$, $y>1$ and $z>1$ then $xyz>1$, which is a contradiction.

By the same way we'll get a contradiction for $x<1$, $y<1$ and $z<1$.

Let $\{a,b\}=\{x,y\}$.

Hence, $(x-1)(y-1)\leq0$ or $x+y\geq xy+1$ and by AM-GM for two numbers we obtain: $$x+y+z\geq xy+z+1\geq2\sqrt{xyz}+1=3$$ and we are done!

Answer 7

The inequality is homogeneous, so it can be assumed WLOG that $z=1\,$, which reduces it to:

$$x+y+1 \ge 3 \cdot \sqrt[3]{xy}$$

But...

• If $\,t \gt 0\,$ then $\,2t^3+1 \ge 3 t^2\,$. This follows from $2t^3-3t^2+1 =(t-1)^2(2t+1) \ge 0\,$

• If $\,x, y \gt 0\,$ then $\,\color{red}{2\sqrt{xy} + 1 \ge 3 \sqrt[3]{xy}}\,$. This follows from the above with $\,t = \sqrt[6]{xy}\,$.

• If $\,x, y \gt 0\,$ then $\,\color{blue}{x+y \ge 2 \sqrt{xy}}\,$ by AM-GM for $\,2\,$ variables, or $\,(\sqrt{x}-\sqrt{y})^2 \ge 0\,$.

Therefore:

$$ x+y+1 \;\color{blue}{\ge}\; 2\sqrt{xy} + 1 \;\color{red}{\ge}\; 3 \sqrt[3]{xy} $$

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[ EDIT ]   See this answer for a similar approach to the general $n \ge 3$ case of AM-GM.