Proof of AM-GM inequality of three terms
$$ \dfrac{a+b+c}{3} \geq \sqrt[3]{abc} $$
$$ (a, b,c 0) $$

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Proof of AM-GM inequality of three terms
$$ \dfrac{a+b+c}{3} \geq \sqrt[3]{abc}   $$
$$ (a, b,c > 0) $$

Uniteasy Admin · Accepted Answer

To prove the AM–GM inequality of three terms
$$ \dfrac{a+b+c}{3} \geq \sqrt[3]{abc}  $$
Let $$ a = x^3, b =y^3, c=z^3 $$
The inequality is converted to 
$$ x^3+y^3+z^3\geq 3xyz $$
Using the identity
$$ x^3+y^3+z^3 - 3xyz $$
$$ = (x+y+z)(x^2+y^2+z^2-xy-yz-xz) $$
$$ =\dfrac{1}{2} (x+y+z)(2x^2+2y^2+2z^2-2xy-2yz-2xz) $$
$$ =\dfrac{1}{2} (x+y+z)[(x-y)^2+(y-z)^2+(x-z)^2] $$
$$ \geq 0 $$
 $$ 	herefore  x^3+y^3+z^3\geq 3xyz $$

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