It is given the function $f:\left( \mathbb{R} - \{0\} \right) \times \left( \mathbb{R}-\{0\} \right) \to \mathbb{R}$ such that $f(a,b)= \left| \frac{|b-a|}{|ab|}+\frac{b+a}{ab}-1 \right|+ \frac{|b-a|}{|ab|}+ \frac{b+a}{ab}+1$ where $a,b \not=0$. Prove that: \[ f(a,b)=4 \cdot \text{max} \left\{\frac{1}{a},\frac{1}{b},\frac{1}{2} \right\}\]
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Tags: function, algebra proposed, algebra
modularmarc101
13.03.2011 17:58
Let $x= \frac{b-a}{ab}$ and $y=\frac{b+a}{ab}$, then $\frac{1}{2}(x+y) = \frac{1}{a}$ and $\frac{1}{2}(y-x) = \frac{1}{b}$. Hence, \[ f(\frac{2}{x+y}, \frac{2}{y-x}) = | |x| + y - 1| + |x| + y + 1 \\ \] and we must prove that \[ | |x| + y - 1| + |x| + y + 1 = 4 \max \{ \frac{x+y}{2}, \frac{y-x}{2}, \frac{1}{2} \} . \\ \]
Notice that $|x| + y - 1 \leq 0 \iff y-1 \leq x \leq 1-y$, so
$\frac{x+y}{2} \leq \frac{(1-y) + y}{2} = \frac{1}{2}$
and
$\frac{y-x}{2} \leq \frac{y - (y-1)}{2} = \frac{1}{2}$,
so $4 \max \{ \frac{x+y}{2}, \frac{y-x}{2}, \frac{1}{2} \} = 4 \cdot \frac{1}{2} = 2$. Also, we have that
$f = -(|x| + y - 1) + |x| + y + 1 = 2$,
so we have proven it for this case.
Now, we must prove it for when $|x| + y - 1 \geq 0$.
We have that $|x| + y - 1 \geq 0 \iff x \leq y-1$ or $x \geq 1-y$. Hence,
$\frac{x+y}{2} \geq \frac{(1-y) + y}{2} = \frac{1}{2}$
and
$\frac{y-x}{2} \geq \frac{y - (y-1)}{2} = \frac{1}{2}$.
Now, if $x \geq 0$, then $\frac{x+y}{2} \geq \frac{y-x}{2}$ and $4 \max \{ \frac{x+y}{2}, \frac{y-x}{2}, \frac{1}{2} \} = 2x+2y$ . Also,
$f = (|x| + y + 1) + |x| + y - 1 = 2x + 2y = 4 \max \{ \frac{x+y}{2}, \frac{y-x}{2}, \frac{1}{2} \} $.
If $x \leq 0$, $\frac{y-x}{2} \geq \frac{x+y}{2}$, so $4 \max \{ \frac{x+y}{2}, \frac{y-x}{2}, \frac{1}{2} \} = 2y-2x$. Also,
$f = (-x + y + 1) - x + y - 1 = 2y - 2y = 4 \max \{ \frac{x+y}{2}, \frac{y-x}{2}, \frac{1}{2} \} $
$\text{Q.E.D.}$