Ineaquality with absolute value

[MathIsEverything]

Prove, that for any real numbers x, y, z satisfying the equality
[math]x^2 + y^2 + z^2 = 1[/math]the ineaquality
[math]|x+y+z| + |x+y-z| + |x-y+z| + |-x+y+z| ≥ 2\sqrt2[/math]is true

 

[Steven G]

Nice problem!
Can we please see your attempt so that we know where you need some help?

 

[Dr.Peterson]

Prove, that for any real numbers x, y, z satisfying the equality
[math]x^2 + y^2 + z^2 = 1[/math]the ineaquality
[math]|x+y+z| + |x+y-z| + |x-y+z| + |-x+y+z| ≥ 2\sqrt2[/math]is true

What axioms or theorems do you have to start from? If this is from a course, what is being taught? If not, what do you know that might be useful?

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[MathIsEverything]

What axioms or theorems do you have to start from? If this is from a course, what is being taught? If not, what do you know that might be useful?

Posting Guidelines (Summary)

Welcome to our tutoring boards! :) This page summarizes the main points from our posting guidelines. As our name implies, we provide math help (primarily to students with homework). We do not generally post immediate answers or step-by-step solutions. We don't do your homework. We prefer to...

www.freemathhelp.com

It’s not from a course, but I was thinking about using AM-GM, but this led me to nowhere and I have no idea how to approach it now. I know when x=y=(sqrt2)/2 and z=0 this expression has the lowest value which is exactly 2*sqrt2 but I don’t know how to prove that there is no scenario where this expression has lower value

 

[stapel]

Note: Duplicate question.