Proof of Triangle Inequality: a+b-c, b+c-a, c+a-b

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SUMMARY

The discussion provides a proof of the Triangle Inequality, specifically demonstrating that for triangle sides a, b, and c, the inequality $\sqrt{a+b-c} + \sqrt{b+c-a} + \sqrt{c+a-b} \leq \sqrt{a} + \sqrt{b} + \sqrt{c}$ holds true. The proof utilizes algebraic manipulation, showing that pairs of square roots can be bounded by twice the square root of the respective side lengths. The steps include deriving inequalities for pairs of sides, ultimately confirming the overall inequality.

PREREQUISITES
  • Understanding of basic algebraic manipulation
  • Familiarity with the properties of square roots
  • Knowledge of triangle properties and inequalities
  • Basic experience with mathematical proofs
NEXT STEPS
  • Study advanced properties of inequalities in geometry
  • Explore the Cauchy-Schwarz inequality and its applications
  • Learn about the Minkowski inequality in mathematical analysis
  • Investigate other proofs of the Triangle Inequality using different methods
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Mathematicians, geometry enthusiasts, students studying inequalities, and anyone interested in mathematical proofs related to triangle properties.

Albert1
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Let a, b, c be the lengths of the sides of a triangle. Prove that:
$\sqrt{a+b-c}$+$\sqrt{b+c-a}$+$\sqrt{c+a-b}\leq\sqrt{a}+\sqrt{b}+\sqrt{c}$
 
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Albert said:
Let a, b, c be the lengths of the sides of a triangle. Prove that:
$\sqrt{a+b-c}$+$\sqrt{b+c-a}$+$\sqrt{c+a-b}\leq\sqrt{a}+\sqrt{b}+\sqrt{c}$

$(\sqrt{a+b-c}$+$\sqrt{b+c-a})^2\leq 4b$
$\therefore \sqrt{a+b-c}$+$\sqrt{b+c-a}\leq 2\sqrt{b}--------------(1)$
likewise
$\therefore \sqrt{b+c-a}$+$\sqrt{c+a-b}\leq 2\sqrt{c}--------------(2)$
$\therefore \sqrt{a+b-c}$+$\sqrt{c+a-b}\leq 2\sqrt{a}--------------(3)$
(1)+(2)+(3) the proof is done
 

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