# Arithmetic Mean-Geometric Mean Inequality

This post is available in PDF. This topic is solely in problem solving in am-gm inequality.  The proof for this inequality can be found here.

Thanks to Engr. Roy Roque Rivera for this very useful article.

Problem 1: Show that if $a,b,c>0$, then $(a+b)(b+c)(c+a) \geq 8abc$

Solution: By AM – GM Inequality, we have

$\displaystyle\frac{a+b}{2}\geq\sqrt{ab}\longleftrightarrow a+b\geq 2\sqrt{ab}$

$\displaystyle\frac{b+c}{2}\geq\sqrt{bc}\longleftrightarrow b+c\geq 2\sqrt{bc}$

$\displaystyle\frac{c+a}{2}\geq\sqrt{ca}\longleftrightarrow c+a\geq 2\sqrt{ca}$

Hence,

$(a+b)(b+c)(c+a) \geq (2\sqrt{ab})(2\sqrt{bc})(2\sqrt{ca})$

or simply

$(a+b)(b+c)(c+a) \geq 8abc$

Problem 2 (AM – HM Inequality): Show that if $x,y>0$, then

$\displaystyle\frac{x+y}{2}\geq\displaystyle\frac{2}{\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}}$

Solution: By AM – GM Inequality, we have

$\displaystyle\frac{\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}}{2}\geq\sqrt{\displaystyle\frac{1}{x}\cdot\displaystyle\frac{1}{y}}$

which can be written as

$\displaystyle\frac{2}{\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}}\leq\sqrt{xy}$

But from AM – GM Inequality, we know that that geometric mean is less than or equal to the arithmetic mean; hence, we obtain
$\displaystyle\frac{x+y}{2}\geq\displaystyle\frac{2}{\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}}$

Comment: In general, for positive real numbers $a_1,a_2,\cdots, a_n$, the following inequality holds
$\displaystyle\frac{a_1+a_2+\cdots+a_n}{n}\geq\root n\of{a_1a_2\cdots a_n}\geq\displaystyle\frac{n}{\displaystyle\frac{1}{a_1}+\displaystyle\frac{1}{a_2}+\cdots\displaystyle\frac{1}{a_n}}$

Problem 3: Show that if $a,b,c>0$, then

$(a+b+c)\displaystyle(\displaystyle\frac{1}{a+b}+\displaystyle\frac{1}{b+c}+\displaystyle\frac{1}{c+a})\geq\displaystyle\frac{9}{2}$

Solution 1:By AM – GM Inequality, we have

$\displaystyle\frac{1}{a+b}+\displaystyle\frac{1}{b+c}+\displaystyle\frac{1}{c+a}\geq 3\root 3\of{\displaystyle\frac{1}{a+b} \cdot \displaystyle\frac{1}{b+c}\cdot\displaystyle\frac{1}{c+a}}$
$(a+b)+(b+c)+(c+a)\geq 3\root 3\of{(a+b)(b+c)(c+a)}$

Hence,

$(a+b)+(b+c)+(c+a)\displaystyle(\displaystyle\frac{1}{a+b}+\displaystyle\frac{1}{b+c}+\displaystyle\frac{1}{c+a})\geq 3\root 3\of{\displaystyle\frac{1}{a+b}\cdot\displaystyle\frac{1}{b+c}\cdot\displaystyle\frac{1}{c+a}}\cdot 3\root 3\of{(a+b)(b+c)(c+a)}$

Simplifying gives

$2(a+b+c)\displaystyle(\displaystyle\frac{1}{a+b}+\displaystyle\frac{1}{b+c}+\displaystyle\frac{1}{c+a})\geq 9$

Thus,

$(a+b+c)(\displaystyle\frac{1}{a+b}+\displaystyle\frac{1}{b+c}+\displaystyle\frac{1}{c+a})\geq\displaystyle\frac{9}{2}$

Solution 2: By AM – HM Inequality, we know that

$\displaystyle\frac{(a+b)+(b+c)+(c+a)}{3}\geq\displaystyle\frac{3}{\displaystyle\frac{1}{a+b}+\displaystyle\frac{1}{b+c}+\displaystyle\frac{1}{c+a}}$

Which can be easily written into

$(a+b+c)\displaystyle(\displaystyle\frac{1}{a+b}+\displaystyle\frac{1}{b+c}+\displaystyle\frac{1}{c+a})\geq\displaystyle\frac{9}{2}$

Problem 4: Show that if $x,y,z>0$, then $x^2+y^2+z^2\geq xy+xz+yz$

Solution 1: By AM – GM Inequality, we have

$x^2+y^2\geq 2xy$ *
$y^2+z^2 \geq 2yz$ **
$z^2+x^2 \geq 2zx$ ***

Adding *,**, and ***, gives us

$2(x^2+y^2+z^2)\geq 2(xy+xz+yz)$

Hence,

$x^2+y^2+z^2\geq xy+xz+yz$

Solution 2: By AM – HM Inequality, we know that

$\displaystyle\frac{(a+b)+(b+c)+(c+a)}{3}\geq\displaystyle\frac{3}{\displaystyle\frac{1}{a+b}+\displaystyle\frac{1}{b+c}+\displaystyle\frac{1}{c+a}}$

Which can be easily written into

$(a+b+c)\displaystyle(\displaystyle\frac{1}{a+b}+\displaystyle\frac{1}{b+c}+\displaystyle\frac{1}{c+a}\geq\displaystyle\frac{9}{2}$

Problem 5: Show that if $x,y,z>0$, then $xy+yz+zx\geq y\sqrt{xz}+z\sqrt{xy}+x\sqrt{zy}$

Solution: By AM – GM Inequality, we have

$xy+yz\geq 2y\sqrt{xz}$ *
$yz+zx \geq 2z\sqrt{xy}$ **
$zx+xy \geq 2x\sqrt{zy}$ ***

Adding *, **, *** gives us

$2(xy+yz+zx)\geq 2(y\sqrt{xz}+z\sqrt{xy}+x\sqrt{zy})$

Hence,

$xy+yz+zx\geq y\sqrt{xz}+z\sqrt{xy}+2x\sqrt{zy}$

More problems to page 2.

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