Triangle Inequality Homework: Show x,y,z Nonnegatives Satisfy Inequality

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Homework Help Overview

The problem involves demonstrating that for nonnegative values of x, y, and z, the inequality (x+y+z)√2 ≤ √(x²+y²) + √(y²+z²) + √(x²+z²) holds true. The context is rooted in the application of the triangle inequality.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the triangle inequality, with one questioning how to relate the given expression to the standard form of the inequality. Others suggest considering the Euclidean norm and its implications for both sides of the equation.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the triangle inequality and its application to the problem. Some have provided insights into the mathematical reasoning behind the norms involved, but no consensus has been reached yet.

Contextual Notes

Participants are working under the constraints of nonnegative values for x, y, and z, and are attempting to clarify the connections between the triangle inequality and the given inequality without providing a complete solution.

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Homework Statement


Show for nonnegative x,y,z that

[tex](x+y+z) \sqrt{2} \leq \sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{x^2+z^2}[/tex]

My book says the answer comes from apply the triangle inequality to (x+y+z,x+y+z)=(x,y)+(y,z)+(x,z). I don't see what they mean by that at all. HOW do you apply the triangle inequality to that?

Homework Equations


The Attempt at a Solution

 
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The standard form of the triangle inequalilty is [itex]d(x,y)\le d(x,z)+ d(y,z)[/itex]. But it is also true that [itex]d(x,z)\le d(z, u)+ d(u, z)[/itex] so, putting those together, you can say that [itex]d(x,y)\le d(x,u)+ d(u,z)+ d(z,y)[/itex]. Do you see to apply that?
 
Take the Euclidian norm of both sides of the equation. For instance, the LHS becomes

||(x+y+z,x+y+z)||=sqrt{(x+y+z)²+(x+y+z)²}=(x+y+z)sqrt{2}
 
quasar987 said:
Take the Euclidian norm of both sides of the equation. For instance, the LHS becomes

||(x+y+z,x+y+z)||=sqrt{(x+y+z)²+(x+y+z)²}=(x+y+z)sqrt{2}

I see. Thanks.
 

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