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.
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.
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!)
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