# Triangle inequality

1. Jul 20, 2008

### ehrenfest

1. The problem statement, all variables and given/known data
Show for nonnegative x,y,z that

$$(x+y+z) \sqrt{2} \leq \sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{x^2+z^2}$$

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?

2. Relevant equations

3. The attempt at a solution

2. Jul 20, 2008

### HallsofIvy

Staff Emeritus
The standard form of the triangle inequalilty is $d(x,y)\le d(x,z)+ d(y,z)$. But it is also true that $d(x,z)\le d(z, u)+ d(u, z)$ so, putting those together, you can say that $d(x,y)\le d(x,u)+ d(u,z)+ d(z,y)$. Do you see to apply that?

3. Jul 20, 2008

### quasar987

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}

4. Jul 20, 2008

### ehrenfest

I see. Thanks.