Suppose $\displaystyle x,y,z$ are any positive real numbers such that $\displaystyle x+y+z=xyz.$ Prove that: $\displaystyle (x-1)(y-1)(z-1) \leq 6\sqrt{3}-10.$

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- Mar 31st 2009, 02:23 AMNonCommAlgPre-algebra: inequality
Suppose $\displaystyle x,y,z$ are any positive real numbers such that $\displaystyle x+y+z=xyz.$ Prove that: $\displaystyle (x-1)(y-1)(z-1) \leq 6\sqrt{3}-10.$

__Source__: JIPAM - Mar 31st 2009, 01:55 PMbkarpuz
We are looking for $\displaystyle \max_{x>0,y>0,z>0}f_{0}(x,y,z),$ where $\displaystyle f_{0}(x,y,z):=(x-1)(y-1)(z-1)$ under the restriction $\displaystyle g(x,y,z):=x+y+z-xyz=0$.

Hence, we set $\displaystyle f_{1}(x,y):=f_{0}(x,y,-(x+y)/(1-xy))$, therefore it suffices to find $\displaystyle \max_{x>0,y>0}f_{1}(x,y)$.

Considering the symmetricity, and it suffices to examine the partial derivative of $\displaystyle f_{1}$ with respect to $\displaystyle x$.

It follows that the critical points are $\displaystyle x_{0}=1$ and $\displaystyle x_{0}=2y/(y^{2}-1)$.

Clearly, $\displaystyle f_{1}(1,y)=0$.

On the other hand, let $\displaystyle f_{2}(y):=f_{1}(2y/(y^{2}-1),y).$

We find that the critical point for $\displaystyle f_{2}$ is $\displaystyle y_{0}=\sqrt{3}$, and $\displaystyle f_{2}(\sqrt{3})=(\sqrt{3}-1)^{3}=6\sqrt{3}-10$, which is the max value of $\displaystyle f_{1}$ (at $\displaystyle (\sqrt{3},\sqrt{3})$) at the same time because the other critical point $\displaystyle (1,y)$ makes $\displaystyle f_{1}$ attain $\displaystyle 0$.

Therefore, the given inequality holds.

**Not**. I know this is a long solution and not so nice. :S - Mar 31st 2009, 04:04 PMNonCommAlg
- Mar 31st 2009, 09:38 PMbkarpuz
(Doh) I know, as I know in JIPAM there are so mant inequalities using convex functions, I guess u wish to see a solution in that direction. But I have no idea since I am not focused on this subject. (Speechless)

I just wanted to share my long and poor solution. (Giggle)