# Thread: Prove (sqr(xy)+sqr(yz)+sqr(xz))^2 >= 3sqr(3)xyz with x + y + z = xyz

1. ## Prove (sqr(xy)+sqr(yz)+sqr(xz))^2 >= 3sqr(3)xyz with x + y + z = xyz

Hi all,

Please help to prove that: (sqr(xy)+sqr(yz)+sqr(xz))^2 >= 3sqr(3)xyz with: x, y, z are positive numbers and x + y + z = xyz.

Thanks.

2. Indication:

$\displaystyle x, y, z > 0 \Rightarrow (\exists )a, b, c \in \left ( -\frac{\pi }{2}, \frac{\pi }{2}\right )$ such that$\displaystyle x=\tan a, y=\tan b, z=\tan c$ (tangent function is surjective)

$\displaystyle x+y+z=xyz\Rightarrow a+b+c=\pi$ (Prove that!)

3. Originally Posted by veileen
Indication:

$\displaystyle x, y, z > 0 \Rightarrow (\exists )a, b, c \in \left ( -\frac{\pi }{2}, \frac{\pi }{2}\right )$ such that$\displaystyle x=\tan a, y=\tan b, z=\tan c$ (tangent function is surjective)

$\displaystyle x+y+z=xyz\Rightarrow a+b+c=\pi$ (Prove that!)
Thanks veileen very much.

If you have another solution without using tan, please show me.

4. Apologies if im missing something, but can you confirm this is the proposition:

$\displaystyle \left(\sqrt{xy} + \sqrt{yz} + \sqrt{xz}\right)^2 \geq 3(\sqrt{3}) xyz$

subject to:
x,y,z >0
xyz = x + y + z.

There are plenty of counter examples, eg:
x=500
y = 1
z = 501/499

(xyz = x+ y + z = 502.004008 as required.)

LHS = 2094.722
RHS = 2608.489

5. Originally Posted by SpringFan25
There are plenty of counter examples,
Agree...
x=1, y=3/2, z=5 :
xyz = x+y+z = 15/2

LHS = 38.432878...
RHS = 38.971143...