# Pre-algebra: inequality

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

Source: JIPAM
• Mar 31st 2009, 01:55 PM
bkarpuz
Quote:

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

Source: JIPAM

We are looking for $\max_{x>0,y>0,z>0}f_{0}(x,y,z),$ where $f_{0}(x,y,z):=(x-1)(y-1)(z-1)$ under the restriction $g(x,y,z):=x+y+z-xyz=0$.
Hence, we set $f_{1}(x,y):=f_{0}(x,y,-(x+y)/(1-xy))$, therefore it suffices to find $\max_{x>0,y>0}f_{1}(x,y)$.
Considering the symmetricity, and it suffices to examine the partial derivative of $f_{1}$ with respect to $x$.
It follows that the critical points are $x_{0}=1$ and $x_{0}=2y/(y^{2}-1)$.
Clearly, $f_{1}(1,y)=0$.
On the other hand, let $f_{2}(y):=f_{1}(2y/(y^{2}-1),y).$
We find that the critical point for $f_{2}$ is $y_{0}=\sqrt{3}$, and $f_{2}(\sqrt{3})=(\sqrt{3}-1)^{3}=6\sqrt{3}-10$, which is the max value of $f_{1}$ (at $(\sqrt{3},\sqrt{3})$) at the same time because the other critical point $(1,y)$ makes $f_{1}$ attain $0$.
Therefore, the given inequality holds.

Not. I know this is a long solution and not so nice. :S
• Mar 31st 2009, 04:04 PM
NonCommAlg
Quote:

Originally Posted by bkarpuz
We are looking for $\max_{x>0,y>0,z>0}f_{0}(x,y,z),$ where $f_{0}(x,y,z):=(x-1)(y-1)(z-1)$ under the restriction $g(x,y,z):=x+y+z-xyz=0$.
Hence, we set $f_{1}(x,y):=f_{0}(x,y,-(x+y)/(1-xy))$, therefore it suffices to find $\max_{x>0,y>0}f_{1}(x,y)$.
Considering the symmetricity, and it suffices to examine the partial derivative of $f_{1}$ with respect to $x$.
It follows that the critical points are $x_{0}=1$ and $x_{0}=2y/(y^{2}-1)$.
Clearly, $f_{1}(1,y)=0$.
On the other hand, let $f_{2}(y):=f_{1}(2y/(y^{2}-1),y).$
We find that the critical point for $f_{2}$ is $y_{0}=\sqrt{3}$, and $f_{2}(\sqrt{3})=(\sqrt{3}-1)^{3}=6\sqrt{3}-10$, which is the max value of $f_{1}$ (at $(\sqrt{3},\sqrt{3})$) at the same time because the other critical point $(1,y)$ makes $f_{1}$ attain $0$.
Therefore, the given inequality holds.

Not. I know this is a long solution and not so nice. :S

well, the question is supposed to be a Pre-algebra (not calculus) problem. that's what the title of my post says. (Evilgrin)
• Mar 31st 2009, 09:38 PM
bkarpuz
Quote:

Originally Posted by NonCommAlg
well, the question is supposed to be a Pre-algebra (not calculus) problem. that's what the title of my post says. (Evilgrin)

(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)