Question

Proof of AM-GM inequality of three terms

\dfrac{a+b+c}{3} \geq \sqrt[3]{abc}

(a, b,c > 0)

Steven Zheng Steven Zheng posted 1 month ago


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

\therefore x^3+y^3+z^3\geq 3xyz

Steven Zheng Steven Zheng posted 1 month ago

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