## Saturday, June 13, 2015

### Given $x,\,y,\,z$ are real such that $2x+y+z+14=2\sqrt{2x}+4\sqrt{y+1}+6\sqrt{z-1}$. Evaluate $\dfrac{x-y}{z}$.

Given $x,\,y,\,z$ are real such that $2x+y+z+14=2\sqrt{2x}+4\sqrt{y+1}+6\sqrt{z-1}$.

Evaluate $\dfrac{x-y}{z}$.

My solution:

This IMO Problem should look easy to you if you're being careful and look at the problem with your heart and mind, not merely with the eyes.

For me, I will try to group

[MATH]\color{yellow}\bbox[5px,purple]{2x,\,2\sqrt{2x}}[/MATH];

[MATH]\color{purple}\bbox[5px,yellow]{y,\,4\sqrt{y+1}}[/MATH];

[MATH]\color{yellow}\bbox[5px,green]{z,\,6\sqrt{z-1}}[/MATH];

so that I have

[MATH]\color{yellow}\bbox[5px,purple]{2x-2\sqrt{2x}}\color{black}+\color{purple}\bbox[5px,yellow]{y-4\sqrt{y+1}}\color{black}+\color{yellow}\bbox[5px,green]{z-6\sqrt{z-1}}\color{black}+14=0[/MATH]

Okay, we have now reached to the point where it's pretty straightforward to complete the square for the following pair:

[MATH]\color{yellow}\bbox[5px,purple]{\begin{align*}2x-2\sqrt{2x}&=2(x-\sqrt{2x})\\&=2\left(\left(\sqrt{x}-\dfrac{\sqrt{2}}{2}\right)^2-\dfrac{2}{4}\right)\\&=2\left(\sqrt{x}-\dfrac{\sqrt{2}}{2}\right)^2-1\end{align*}}[/MATH]

But it's not so simple to complete the square for [MATH]\color{purple}\bbox[5px,yellow]{y,\,4\sqrt{y+1}}[/MATH] and [MATH]\color{yellow}\bbox[5px,green]{z,\,6\sqrt{z-1}}[/MATH].

Yes, I said it's not very simple, but it's entirely possible if we borrow a [MATH]\color{purple}\bbox[5px,yellow]{+1}\color{black}\,\,\text{for}\color{purple}\bbox[5px,yellow]{y}[/MATH] and minus another [MATH]\color{yellow}\bbox[5px,green]{-1}\color{black}\,\,\text{for}\color{yellow}\bbox[5px,green]{z}[/MATH], we see that we have:

[MATH]\color{purple}\bbox[5px,yellow]{\begin{align*}y+1-4\sqrt{y+1}&=(y+1)-4\sqrt{y+1}\\&=(\sqrt{y+1}-2)^2-4\end{align*}}[/MATH]

[MATH]\color{yellow}\bbox[5px,green]{\begin{align*}z-1-6\sqrt{z-1}&=(z-1)-6\sqrt{y+1}\\&=(\sqrt{z-1}-3)^2-9\end{align*}}[/MATH]

See, we can complete the square for the other two pairs if we performed some algebraic tricks. :D

Now, if we do this to the given original equation, we get:

$2x+y+z+14=2\sqrt{2x}+4\sqrt{y+1}+6\sqrt{z-1}$

$2x-2\sqrt{2x}+y-4\sqrt{y+1}+z-6\sqrt{z-1}+14=0$

$(2x-2\sqrt{2x})+(y+1-4\sqrt{y+1}-1)+(z-1-6\sqrt{z-1}+1)+14=0$

$(2x-2\sqrt{2x})+(y+1-4\sqrt{y+1}-\cancel{1})+(z-1-6\sqrt{z-1}+\cancel{1})+14=0$

$(2x-2\sqrt{2x})+(y+1-4\sqrt{y+1})+(z-1-6\sqrt{z-1})+14=0$

$\left(2\left(\sqrt{x}-\dfrac{\sqrt{2}}{2}\right)^2-1\right)+(\sqrt{y+1}-2)^2-4+(\sqrt{z-1}-3)^2-9+14=0$

$2\left(\sqrt{x}-\dfrac{\sqrt{2}}{2}\right)^2+(\sqrt{y+1}-2)^2+(\sqrt{z-1}-3)^2-1-4-9+14=0$

$2\left(\sqrt{x}-\dfrac{\sqrt{2}}{2}\right)^2+(\sqrt{y+1}-2)^2+(\sqrt{z-1}-3)^2-14+14=0$

$2\left(\sqrt{x}-\dfrac{\sqrt{2}}{2}\right)^2+(\sqrt{y+1}-2)^2+(\sqrt{z-1}-3)^2=0$

We can see it clearly now that the equation above is the sum of three square terms. The sum of the terms that are always greater than or equal to zero will either be

1. zero when each of them is zero, or

2. always greater than zero.

What does that mean literally?

That means each of the square terms must equal zero, that is the only way the LHS can be set to take the zero value.

That is, we have:

$2\left(\sqrt{x}-\dfrac{\sqrt{2}}{2}\right)^2=0\,\,\implies\,\,\sqrt{x}=\dfrac{\sqrt{2}}{2}$, i.e. $x=\dfrac{1}{2}$

$(\sqrt{y+1}-2)^2=0\,\,\implies\,\,\sqrt{y+1}=2$, i.e. $y=4-1=3$

$(\sqrt{z-1}-3)^2=0\,\,\implies\,\,\sqrt{z-1}=3$, i.e. $z=9+1=10$

Therefore, $\dfrac{x-y}{z}=\dfrac{\dfrac{1}{2}-3}{10}=-\dfrac{1}{4}$.