Find minimal value of $a \in \mathbb{R}$ such that system $$\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}=a-1$$$$\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1}=a+1$$has solution in set of real numbers
Problem
Source: Regional Olympiad - Federation of Bosnia and Herzegovina 2009
Tags: System, algebra, minimum, inequalities
28.09.2018 17:56
gobathegreat wrote: Find minimal value of $a \in \mathbb{R}$ such that system $$\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}=a-1$$$$\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1}=a+1$$has solution in set of real numbers Let $\Delta_x=\sqrt{x+1}-\sqrt{x-1}=\frac 2{\sqrt{x+1}+\sqrt{x-1}}\in(0,\sqrt 2]$ We are looking for $\min(\frac 1{\Delta_x}+\frac 1{\Delta_y}+\frac 1{\Delta_z})$ with conditions $\Delta_i\in(0,\sqrt 2]$ and $\sum\Delta_i=2$ And it is easy, with this form, to establish that $\min\sum\frac 1{\Delta_i}$ is reached when $\Delta_x=\Delta_y=\Delta_z=\frac 23$ And so is $\boxed{\frac 92}$
29.09.2018 12:15
gobathegreat wrote: Find minimal value of $a \in \mathbb{R}$ such that system $$\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}=a-1$$$$\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1}=a+1$$has solution in set of real numbers Solution of dragonheart6:
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29.09.2018 12:34
gobathegreat wrote: Find minimal value of $a \in \mathbb{R}$ such that system $$\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}=a-1$$$$\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1}=a+1$$has solution in set of real numbers Solution of Zhangyunhua: $$a=\frac{1}{2}\sum_{cyc}(\sqrt{x+1}+\sqrt{x-1})=\sum_{cyc}\frac{1}{\sqrt{x+1}-\sqrt{x-1}}\geq\frac{9}{\sum_{cyc}\sqrt{x+1}-\sum_{cyc}\sqrt{x-1}}=\frac{9}{2}.$$Equality holds when $x=y=z=\frac{85}{36}.$
02.04.2020 23:13
Here is nice solution. We have system of equations and let us subtract second equation with first. $(\sqrt{x+1}-\sqrt{x-1})+(\sqrt{y+1}-\sqrt{y-1})+(\sqrt{z+1}-\sqrt{z-1})=2$, now we can use that: $\sqrt{k+1}-\sqrt{k-1}=\frac{2}{\sqrt{k+1}+\sqrt{k-1}}$ (easy to prove), we can replace $k$ with $x, y, z$ and get this. $\frac{2}{\sqrt{x+1}+\sqrt{x-1}}+\frac{2}{\sqrt{y+1}+\sqrt{y-1}}+\frac{2}{\sqrt{z+1}+\sqrt{z-1}}=2$ $\frac{1}{\sqrt{x+1}+\sqrt{x-1}}+\frac{1}{\sqrt{y+1}+\sqrt{y-1}}+\frac{1}{\sqrt{z+1}+\sqrt{z-1}}=1$, now we use $AM-HM$ inequality we get that $1 \geq \frac{9}{\sqrt{x+1}+\sqrt{x-1}+\sqrt{y+1}+\sqrt{y-1}+\sqrt{z+1}+\sqrt{z-1}}=\frac{9}{2a}$ or $a \geq \frac{9}{2}$ Equality is when $a=\frac{9}{2}$ $\implies$ $\sqrt{x-1}+\sqrt{x+1}=\sqrt{y-1}+\sqrt{y+1}=\sqrt{z-1}+\sqrt{z+1}=3$, and now just solve simple equation: $\sqrt{x-1}+\sqrt{x+1}=3$ or $x-1+2\sqrt{x^2-1}+x+1=9$ $\implies$ $4x^2-4=81-36x+4x^2$, and finally $x=y=z=\frac{85}{36}$ That is all $Q.E.D.$