hello
I have something that i can't solve.
There is a function f:R3-R with f(x,y,z)=xy+xz+yz and i know that xyz=1, how do i prove that f(x,y,z)f(1,1,1)=3 without just replacing the values 1,1,1 ?

Question

hello
I have something that i can't solve.
There is a function f:R3->R with f(x,y,z)=xy+xz+yz and i know that xyz=1, how do i prove that f(x,y,z)>f(1,1,1)=3 without just replacing the values 1,1,1 ?

anonymous · Answer

divide f(x,y,z)by xyz

anonymous · Answer

i devided again and it's 1/z+1/y+1/z .... but i m not allowed to replace 1,1,1 :) SO WHAT NEXT ?

anonymous · Answer

I'm going to type this up on a LaTeX pdf as it's too long to write up on here, comfortably.

anonymous · Answer

thank u wolf :D

anonymous · Answer

What class is this for? I'm coming up with an elementary proof, but if this is for calc or something, maybe I could use something slightly more sledge-hammer-ish? It's just a bunch of cases and lemmas.

anonymous · Answer

Im confused by R3->R...

anonymous · Answer

It maps the vector-space $\mathbb{R}^3$ into the real space.

anonymous · Answer

the function is defined in the space R3 with values in R

anonymous · Answer

well i solved it with Lagrange multiplicators but the prof. asked for elementary proof :)) which beats me

anonymous · Answer

Oh, it's not too difficult, it's just kind of tedious... but nice... lol.

anonymous · Answer

it's not nice for me :P

anonymous · Answer

And, for the last proof, note that:
$$
(y-z)^2>zy
$$Working back up from there gets you the desired result.

anonymous · Answer

There's one more case to consider, where x<1, yz>1, but it's the same as the other one, if you solve it generally.

anonymous · Answer

yeah :D figured it out :D thank u a lot :D

anonymous · Answer

Wait, wait, I believe there might be a mistake in my last statement "$(y-z)^2>zy$"... agh.