Prove Triangle Inequality: AB/MZ + AC/ME + BC/MD ≥ 2t/r

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The discussion centers on proving the triangle inequality expressed as $$\frac{AB}{MZ}+\frac{AC}{ME}+\frac{BC}{MD}\geq\frac{2t}{r}$$, where t represents half the perimeter of triangle ABC and r denotes the radius of the inscribed circle. The proof involves the application of the Cauchy-Schwarz inequality, which is essential for establishing the relationship between the sides of the triangle and the perpendiculars drawn from point M. This inequality is a fundamental concept in geometry and is critical for solving problems related to triangle properties.

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solakis1
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Given a triangle ABC and a point M inside the triangle ,draw perpendiculars MZ,MD,ME at the sides AB,BC,AC respectively. Then prove:$$\frac{AB}{MZ}+\frac{AC}{ME}+\frac{BC}{MD}\geq\frac{2t}{r}$$

Where t is half the perimeter of the triangle and r is the radius of the inscribed circle
 
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[sp]Here also the Cauchy-Schwarz inequality may be used for the solution of the problem[/sp]
 

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