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

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The discussion focuses on proving the inequality involving a triangle ABC and a point M inside it, where perpendiculars MZ, MD, and ME are drawn to the sides AB, BC, and AC, respectively. The goal is to demonstrate that the sum of the ratios of the triangle's sides to these perpendiculars is greater than or equal to twice the semi-perimeter divided by the inradius. The Cauchy-Schwarz inequality is suggested as a useful tool for the proof. Participants explore various approaches and mathematical principles to establish the validity of the inequality. The proof highlights the relationship between triangle geometry and inequalities, emphasizing the significance of the inscribed circle's radius.
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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]
 
Thread 'Erroneously finding discrepancy in transpose rule'

Thread 'Erroneously finding discrepancy in transpose rule'

Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...
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