The problem:(adsbygoogle = window.adsbygoogle || []).push({});

"[tex]xy + yz + zx = 12[/tex]

[tex]xyz = 2 + x + y + z[/tex]

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;

[tex]\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{12}{2 + x + y + z}[/tex]

or

[tex]2 + x + y + z = \frac{12}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}[/tex]

if we do this:

[tex]\frac{2 + x + y + z}{4} = \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}[/tex]

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.

EDIT:

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.

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# Simultaneous Equations

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