Multivariable Calc question, please help!

Let x, y, z be positive real numbers. Prove that:

sqrt(2) (x+y+z)

Question

Multivariable Calc question, please help!

Let x, y, z be positive real numbers. Prove that:

sqrt(2) (x+y+z) <= sqrt(x^2+y^2) + sqrt(y^2+z^2) + sqrt(z^2+x^2)

anonymous · Answer

i'm thinking it has something to do with (x+y+z)^2 ≥ 0 and somehow manipulate the algebra

anonymous · Answer

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

anonymous · Answer

and x^2 + y^2 + z^2 ≥ 0

anonymous · Answer

should i square both sides? btw thank you i really appreciate it

anonymous · Answer

no, you don't assume the conclusion because that's what you're supposed to show

anonymous · Answer

(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2xz + 2yz ≥ 0

huhm... what can we do from here?

anonymous · Answer

what happened to the sqrt(2) that was on the LHS?

anonymous · Answer

well that's what i'm trying to figure out :D

anonymous · Answer

lol haha

anonymous · Answer

do u have an idea as to what this may be called? is there a name for a proof like this?

anonymous · Answer

no :D

anonymous · Answer

Unfortunately, I am not very good at Mathematics. Much less Calculus. :P

anonymous · Answer

@linda3 is good at this though I think :P

anonymous · Answer

Thanks @bookworm00981 
Hey @linda3 do you think you'll be able to help out?

linda3 · Answer

yeah

linda3 · Answer

no not really my brain is a bit fried right now... I just finished a Calculus Test :P but... here @satellite73 , @Compassionate , @Luigi0210

linda3 · Answer

i'm sorry but my friends can probably help you

anonymous · Answer

Lol :P

luigi0210 · Answer

@ganeshie8 @Loser66 @austinL

anonymous · Answer

this is killing me! lol what part of calculus is this so i can look it up somewhere? I just need some hints "/