Inequality

Let x, y,z >0 satisfying x+y+z=1. Prove that

    \[\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \sqrt{\frac{3}{xyz}}\]

Solution

Since \frac{1}{x} \; , \; \frac{1}{y} \; , \; \frac{1}{z} >0 then the numbers

    \[\sqrt{\frac{1}{x} + \frac{1}{y}} \; , \; \sqrt{\frac{1}{x} +\frac{1}{z}} \; , \; \sqrt{\frac{1}{y} + \frac{1}{z}}\]

could be sides of a triangle. The area of this triangle is

    \[\mathcal{A} = \frac{1}{2} \sqrt{\frac{1}{xy} + \frac{1}{xz} + \frac{1}{yz}} = \frac{1}{2} \sqrt{\frac{x+z+y}{xyz}} = \frac{1}{2\sqrt{xyz}}\]

However , in any triangle is holds that [Weitzenböck]

(1)   \begin{equation*} a^2+b^2+c^2 \geq 4 \mathcal{A} \sqrt{3}  \end{equation*}

where \mathcal{A} is the area of the triangle. Thus

    \[\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \sqrt{\frac{3}{xyz}}\]

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