# Prove $xy+yz+zx \geq 9xyz$

If $$x,y,z>0$$ and $$x+y+z=1$$, prove $$xy+yz+zx \geq 9xyz$$.

Can anyone help me? I think I'm close, but I can't get to the end. Using geometric and arithmetic inequality, I got $$(xy+yz+xz)(2+xyz) \geq 9xyz$$.

Thank you

• It might be helpful for others if you explained in more detail how you got as far as you did. – Robert Shore Oct 14 at 21:55

We have that

$$xy+yz+zx \geq 9xyz \iff \frac1x+ \frac1y+\frac1z\ge 9$$

and by HM-AM inequality

$$\frac3{\frac1x+ \frac1y+\frac1z}\le \frac{x+y+z}3$$

Alternatively, \begin{aligned} xy+yz+zx &= (x+y+z)(xy+yz+zx) \\ &\ge 3(x\cdot y\cdot z)^{1/3}\cdot 3(xy\cdot yz\cdot zx)^{1/3} \\ &=9xyz\ . \end{aligned}