MHB How to Solve for Triples in Equation with Absolute Value?

[anemone]

Here is this week's POTW:

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Find all triples $(a,\,b,\,c)$ of real numbers that satisfy $a^2+b^2+c^2+1=ab+bc+ca+|a-2b+c|$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

No one answered last week's problem.(Sadface)

You can find the suggested solution as follows:

Spoiler

We rewrite the given equation as

$\dfrac{a^2}{2}-ab+\dfrac{b^2}{2}+\dfrac{b^2}{2}-bc+\dfrac{c^2}{2}+\dfrac{c^2}{2}-ca+\dfrac{a^2}{2}+1=|a-2b+c|$

Or equivalently,

$\dfrac{(a-b)^2}{2}+\dfrac{(b-c)^2}{2}+\dfrac{(c-a)^2}{2}1=|a-b+c-b|$*

Substitute $x=a-b$ and $y=c-b$, we get $a=x-y$, thus we have

$\dfrac{x^2}{2}+\dfrac{y^2}{2}+\dfrac{(x-y)^2}{2}1=|x+y|$**

From $(x-y)^2\ge 0$, it follows that $x^2-2xy+y^2\ge 0$, hence $2x^2+2y^2\ge x^2+y^2+2xy$, which means that $x^2+y^2\ge \dfrac{x+y)^2}{2}$, with equality if and only if $x=y$. Furthermore, note that $(x-y)^2\ge 0$. Hence we get

$\begin{align*}|x+y|&=\dfrac{x^2}{2}+\dfrac{y^2}{2}+\dfrac{(x-y)^2}{2}1\\&\ge \dfrac{(x+y)^2}{4}\end{align*}$

Now write $z=|x+y|$, then the expression becomes $z\ge \dfrac{z^2}{4}+1$, which actually is $z^2-4z+4=(z-2)^2\le 0$.

Since the LHS is a square, equality must hold, so $z=2$. Furthermore, in our previous inequalities, equality also has to hold so $x=y$. Substituting this in equation ** gives $x^2+1=2$, so $x=\pm 1$. Thus we find the triples $(b+1,\,b,\,b+1)$ and $(b-1,\,b,\,b-1)$ for arbitrary $b\in \Bbb{R}$.

Substituting this in equation * shows that these triples are indeed solutions for all $b\in \Bbb{R}$.