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Difference between revisions of "ASIA TEAM Problems/Problem 6"

(Created page with "Dividing both sides of the second equation by <math>xyz</math> gives <math>\frac{9}{x}+\frac{4}{y}+\frac{1}{z}=6</math>. But the Cauchy-Swarsz inequality gives <math>(x+y+z)(\...")
 
(No difference)

Latest revision as of 04:43, 16 February 2024

Dividing both sides of the second equation by $xyz$ gives $\frac{9}{x}+\frac{4}{y}+\frac{1}{z}=6$. But the Cauchy-Swarsz inequality gives $(x+y+z)(\frac{9}{x}+\frac{4}{y}+\frac{1}{z})$ $\geq(\sqrt{x\times\frac{9}{x}}+\sqrt{y\times\frac{4}{y}} +\sqrt{z\times\frac{1}{z}})^2$ $=(3+2+1)^2=36.$ Since $(x+y+z)(\frac{9}{x}+\frac{4}{y}+\frac{1}{z})=6\times6=36$, we have the equality case, i.e. $x:y:z=\frac{9}{x}:\frac{4}{y}:\frac{1}{z}$, or $x:y:z=3:2:1$, or $(x,y,z)=(3,2,1)$, and $xyz=\boxed{6}$. ~AbbyWong

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