Proving the Triangle Inequality in Real Analysis: abs(abs(x)-abs(y))<=abs(x-y)

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SUMMARY

The discussion focuses on proving the triangle inequality in real analysis, specifically the inequality abs(abs(x)-abs(y)) <= abs(x-y). The participants utilize the triangle inequality property abs(a+b) <= abs(a) + abs(b) to approach the proof. The initial attempt involves substituting a = x - y and b = y, leading to the expression abs(x) - abs(y) <= abs(x-y). The discussion highlights the need to demonstrate the reverse inequality |y| - |x| <= |x-y| for a complete proof.

PREREQUISITES
  • Understanding of real analysis concepts, particularly the triangle inequality.
  • Familiarity with absolute value properties in mathematics.
  • Basic algebraic manipulation skills.
  • Knowledge of inequalities and their proofs in mathematical contexts.
NEXT STEPS
  • Study the properties of absolute values in real analysis.
  • Learn about the triangle inequality and its applications in proofs.
  • Explore alternative proofs of the triangle inequality.
  • Practice solving inequalities involving absolute values.
USEFUL FOR

Students studying real analysis, mathematics educators, and anyone interested in understanding the proofs of inequalities involving absolute values.

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Homework Statement



Prove: abs(abs(x)-abs(y))<=abs(x-y)

Homework Equations



Triangle Inequality:
abs(a+b)<=abs(a)+abs(b)

The Attempt at a Solution



This is what i have so far:

Let a=x-y and b=y. Then
abs(x-y+y) <= abs(x-y)+abs(y) which becomes abs(x)-abs(y)<=abs(x-y). From here i get stuck can anybody help me?
 
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Well you've got half of it. The other inequality you need to demonstrate is |y| - |x| <= |x-y|. But what is an equivalent way of writing |x-y|?
 

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