Question

Given x,y,z >0 such that \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1, find the minimum value of \dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}

Collected in the board: Inequality

Steven Zheng posted 4 months ago

Answer

Using the AM-GM inequality,

\dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}

=\dfrac{1}{2}( \dfrac{x}{yz}+\dfrac{y}{xz})+\dfrac{1}{2}(\dfrac{y}{xz}+\dfrac{z}{xy}) +\dfrac{1}{2}(\dfrac{z}{xy}+\dfrac{x}{yz})

\geq \sqrt{\dfrac{x}{yz}\cdotp \dfrac{y}{xz} } +\sqrt{\dfrac{y}{xz}\cdotp \dfrac{z}{xy} } +\sqrt{\dfrac{z}{xy}\cdotp \dfrac{x}{yz} }

=\sqrt{\dfrac{1}{z^2 } }+\sqrt{\dfrac{1}{x^2} }+\sqrt{\dfrac{1}{y^2} }

=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}

=1

So the minimum value of \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} is 1 when x=y=3


Steven Zheng posted 4 months ago

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