# Prove that $\dfrac {2x+y+z}{x}+\dfrac {2y+x+z}{y}+\dfrac {2z+x+y}{z}\ge 12$

I am trying to prove that

$$\dfrac {2x+y+z}{x}+\dfrac {2y+x+z}{y}+\dfrac {2z+x+y}{z}\ge 12$$

Clearly

$$2x+y+z>x, \ 2y+x+z>y, 2z+x+y>z$$

However, I struggle to find a suitable inequality relation to move further on here. Does anyone have a suggestion?

Thanks

• You should tell us where $x,y,z$ live--are they real numbers? Also, please show any work you've initially done. Commented Aug 21, 2023 at 12:35
• That is right, $x,\ y,\ z \in \mathbb{N}$ Commented Aug 21, 2023 at 12:36
• Hint: the LHS is related to $(x+y+z)(1/x + 1/y + 1/z)$. Commented Aug 21, 2023 at 12:36
• @NickF Now we can re-write this to $$\dfrac{(x^2 y + x^2 z + x y^2 + 6 x y z + x z^2 + y^2 z + y z^2)}{x y z}\geq 12$$ and since we have the following equality: $$(x+y+z)(1/x+1/y+1/z)=\frac{(x + y + z) (x y + x z + y z)}{x y z}$$ We can further develop it. But how do we go on from there? Commented Aug 21, 2023 at 12:51

LHS = $$3 + (x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \ge 3 + 9 = 12$$ by AM-HM inequality.
• $\frac{2x+y+z}{x} = 1 + \frac{x+y+z}{x}$. Can you take it from there? Commented Aug 21, 2023 at 13:02
• No, do the same thing to each of the three terms, that's where the 3 comes from and you are left with $(x+y+z)(1/x+1/y+1/z)$. Commented Aug 21, 2023 at 13:12