When does the triangle inequality hold for absolute value?

lepton123
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Homework Statement


abs(x+y+z)≤abs(x)+abs(y)+abs(z) indicate when this equality holds and prove this statement


Homework Equations



Triangle inequality?

The Attempt at a Solution


I have nothing :/
 
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What is the absolute value of x? if x => 0 then abs(x)=x, else it is -x.

So make a table with all of the possible cases and see what happens!
 
Drat, I was hoping that that I wouldn't have to do a case analysis; is there a more elegant way of solving this though?
 
I don't know elegant ... I grew up on a farm!

So once you have carried out the detailed work you can apply your own standards of elegance and cleverness ... and write something elegant!
 
Assume (by renaming of variables) that x ≤ y ≤ z. Then you have 4 cases to check, it shouldn't be too tedious.
 
x + y + z <= |x| + |y| + |z|
- x - y - z <= |x| + |y| + |z|

(definition of the absolute value)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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