Prove x+y+z <= x^2+y^2+z^2 where?

Question

where

x, y, z all > 0
and
xyz =1

I've done similar problems in the past. I know that I can shift everthing to one side and put in a factored form:

0 <= (x-1/2)^2+(y-1/2)^2+(z-1/2)^2 -3/4

but where to go from here, or even what to do.

Looking at the conditions, since x,y, and z > 0 then we can express
x=1/(yz)

and similar with y, z...

Any ideas?

To JJ:

Huh? What about x=1, y=2, z=1/2

then xyz=1

I tried to go back and tone down what I had said, JJ, but found out that I could only add more details, not edit details. Oh well. What you did is the kind of thing that one does when taking a quick look and quick jab at solving.

Thanks for you effort. Perhaps you can return tomorrow and see what's been posted. You'll probably lose sleep thinking about it!

Thanks again, JJ.

You don't seem to want to allow any numbers between 0 and 1.

x<1 -> x > x^2
x>1 -> x < x^2

Gianlino;
Thanks for the effort. However,
x,y,z are restricted to > 0 not just >1. Your contribution is somewhat complicated and it's late, so I'll give it more attention after some snoozing.

Answer 1

ok, so start with x, if x is an element of positive real numbers, x is less than x^2. this is a fact. do the same for y and z, then add the equations. im too lazy to go through all the steps. its pretty simple, i think. lol, sry for the other answer, i was thinking the characteristics of a zero product...

Answer 2

If x, y , and z=1, non problem. So you may assume z>1. For a given value of (x+y) the minimum value of (x^2 + y^2) will be (x+y)^2 / 2. So it is enough to show that x+y+z < (x+y)^2/2 + z^2.
So now you just want to show u+z u is the sum of two numbers x and y such that xy=1/z , that is u^2-4/z >=0. The minimum of u^2/2 -u + z^2 - z as a function of u, is attained for u=1 and is z^2-z-1/2. So if z> 2 there is no problem. If z< 2 remember that u^2>4/z. So the minimum has to be considered only on the interval u>= 2/sqrt{z} where it is
2/z - 2/sqrt{z} + z^2 -z. Letting w=sqrt {z} you get to study the sign of 2 -2w + w^6 - w^4 for 1

Answer 3

i dont know the answer, but the above answer is incorrect. If xyz=1 it doesnt mean x=1, y=1, z=1. What if x=1, y=2, and z=.5?

x, y, z > 0 gives you an easy out. Also your factored form is wrong. This might be better:

0<= x(x-1)+y(y-1)+z(z-1)

Answer 4

method to prove question like these is here: You surly know that (x-y)^2 => 0 so: x^2 + y^2 - 2xy => 0 ==> x^2 + y^2 => 2xy for 3xyz <= x^3 + y^3 + z^3 and your question do like this.

Answer 5

You're on the right track. You need to show that

x² + y² + z² - x - y - z > 0

You started out with the right idea--factoring.
x² + y² + z² - x - y - z
= (x - ½)² + (y - ½)² + (z - ½)² - ¾
We added in ¾ in the factoring so we need to take it out again.

Hmm...I need to think about the rest.