Yay for inequality
by shiningsunnyday, Apr 14, 2017, 9:10 AM
2013 USAMO 4 wrote:
Find all real numbers
satisfying ![\[\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.\]](//latex.artofproblemsolving.com/2/6/e/26edbd87f49eb68ec9b662b602db712e5151dc9c.png)

![\[\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.\]](http://latex.artofproblemsolving.com/2/6/e/26edbd87f49eb68ec9b662b602db712e5151dc9c.png)
Solution
We will make use of Cauchy of the form
WLOG let
so that
, and from Cauchy,
In order for the original equation to be true, equality must be present in both inequalities, giving us ![\[x-1 = \frac{1}{yz}\]](//latex.artofproblemsolving.com/d/3/b/d3b3cd562d28b9398742a479b5d56c84561d394d.png)
which gives us the triplet
for 
However, we have to check our assumption:![\[y \ge x \Rightarrow y \ge \frac{y^2+y-1}{y^2} \Rightarrow y^3 -y^2-y+1 = (y+1)(y-1)^2 \ge 0\]](//latex.artofproblemsolving.com/7/4/6/7468b2a9d10d21dd1938e77aabb8cbb1fcf2e174.png)
Also,
as otherwise that would make
undefined. Specifically, the original condition breaks down to
but as we previously had gotten
contradiction. Finally, we check that ![\[\frac{y}{y-1} = 1 + \frac{1}{y-1} > 1\]](//latex.artofproblemsolving.com/5/8/2/582df949d003d20ef57befe9a550e354f748063b.png)
Therefore, we conclude
for
generates our solution set.
![\[a_1b_1 + a_2b_2 + \ldots + a_n b_n \le \sqrt{a_1^2+a_2^2+ \ldots + a_n^2} \cdot \sqrt{b_1^2+b_2^2+ \ldots + b_n^2}.\]](http://latex.artofproblemsolving.com/5/4/f/54fd6110e36d0f1a2fc5ba503c59c41f0d2b88cd.png)

![\[\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x+xyz} = \sqrt{(x-1+1)(1+yz)}\]](http://latex.artofproblemsolving.com/5/2/a/52a2b979127ca95766c2404e3380546c317ae1c8.png)
![\[\sqrt{(x-1+1)(1+yz)} \ge \sqrt{x-1}+\sqrt{yz}=\sqrt{x-1}+\sqrt{(y-1 +1)(1+ z-1)} \ge \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.\]](http://latex.artofproblemsolving.com/d/7/a/d7a084b72655f0eafc291733da4e2758f05ceeea.png)
![\[x-1 = \frac{1}{yz}\]](http://latex.artofproblemsolving.com/d/3/b/d3b3cd562d28b9398742a479b5d56c84561d394d.png)
![\[y-1 = \frac{1}{z-1},\]](http://latex.artofproblemsolving.com/1/3/c/13c1c6cfa9e0f7b8e2ae8062a216e36ce14153b5.png)
![\[(x, y, z) = \left(\frac{y^2+y-1}{y^2}, y, \frac{y}{y-1} \right)\]](http://latex.artofproblemsolving.com/3/5/c/35c03ce49e20f2eef7a81031dea41728542052d9.png)

However, we have to check our assumption:
![\[y \ge x \Rightarrow y \ge \frac{y^2+y-1}{y^2} \Rightarrow y^3 -y^2-y+1 = (y+1)(y-1)^2 \ge 0\]](http://latex.artofproblemsolving.com/7/4/6/7468b2a9d10d21dd1938e77aabb8cbb1fcf2e174.png)
![\[z \ge x \Rightarrow \frac{y}{y-1} \ge \frac{y^2+y-1}{y^2} \Rightarrow y^3 \ge y^3-2y+1 \Rightarrow 2y \ge 1.\]](http://latex.artofproblemsolving.com/c/a/5/ca5f502f389874abbe54c7fe899c78db5a177375.png)


![\[\sqrt{1+zx} = \sqrt{x-1}+\sqrt{z-1}\]](http://latex.artofproblemsolving.com/6/a/5/6a50a0a4977c1de1a5330dff7a5dbbf70e6feaed.png)
![\[\sqrt{1+zx} > \sqrt{zx} = \sqrt{(z-1+1)(1+x-1)} \ge \sqrt{z-1}+\sqrt{x-1},\]](http://latex.artofproblemsolving.com/9/d/d/9dd08f065e62a2add79fe682b9542c2bf5aaa52a.png)
![\[\frac{y}{y-1} = 1 + \frac{1}{y-1} > 1\]](http://latex.artofproblemsolving.com/5/8/2/582df949d003d20ef57befe9a550e354f748063b.png)
![\[\frac{y^2+y-1}{y^2} = 1 + \frac{y-1}{y^2} > 1.\]](http://latex.artofproblemsolving.com/f/a/1/fa123f6d78ac187d316df4f5b2a9630b3a708415.png)


Tidbit
Hm so 2013 1/4 took me collectively ~50 mins, which I guess is a good sign?