or

so

and since x, y, z > 0 (all positive), then xyz > 2, right?

But, again, this looks pointless, seeing as xyz = 2 + x + y + z, so of course xyz > 2.

Can anyone give me a push in the right direction for this? This question seems to look like it is designed to use AM-GM but I am not seeing what kind of upper or lower bound I can get on xyz using it.

]]>Oh, that is it! I was assuming only integers.

]]>I posted the exact wording of the problem as shown on my problem sheet -- I am assuming by 'distinct' they mean x, y and z must have different values to each other.

]]>"

Find a solution of the above simultaneous equations, in which all of x, y and z are positive, and prove that it is the only such solution.

Show that a solution exists in which x, y and z are real and distinct."

I haven't really made much progress on this problem. I divided the first equation by the second and got;

or

if we do this:

I notice that the LHS is the arithmetic mean of {2, x, y, z} and the RHS is the harmonic mean of {x, y, z}. They are equal iff x = y = z = 2, so one solution is x = 2, y = 2, z = 2. BUT it appears that they want x, y and z to be distinct.

Can anyone help me here?

Thanks.

Wait... on second thought, what I wrote is wrong. The two sets {2,x,y,z} and {x,y,z} aren't identical. Looks like I am back to square one.

]]>