AM-GM Inequality for Three Variables

Inequality of Arithmetic and Geometric Means (AM-GM) states that for all positive real numbers $x_1$, $x_2$, $x_3$ it holds that

$$ \frac{ x_1+x_2+x_3 }{3} \geq \sqrt[3]{ x_1x_2x_3 } $$

The equality holds if $x_1=x_2=x_3$.

Problem (Moscow City, 2008)

Let $x$, $y$, $z$ be positive real numbers, such that $xyz=1$. Show that

$$ (2+x)(2+y)(2+z) \geq 27 $$

 

Solution

Let us apply the AM-GM inequality to the numbers $1$, $1$ and $t$:

$$ 1+1+t \geq 3 \sqrt[3]{t} $$

Therefore we have

$$ 2+x = 1+1+x \geq 3 \sqrt[3]{x} $$

$$ 2+y = 1+1+y \geq 3 \sqrt[3]{y} $$

$$ 2+z = 1+1+z \geq 3 \sqrt[3]{z} $$

 

and thus

$$ (2+x)(2+y)(2+z) \geq 3 \sqrt[3]{x} \cdot 3 \sqrt[3]{y} \cdot 3 \sqrt[3]{z} = 27 $$

which is what needed to be proven.

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