Let $a,b,c$ be reals and \[f(a, b, c) = \left| \frac{ |b-a|}{|ab|} +\frac{b+a}{ab} -\frac 2c \right| +\frac{ |b-a|}{|ab|} +\frac{b+a}{ab} +\frac 2c\] Prove that $f(a, b, c) = 4 \max \{\frac 1a, \frac 1b,\frac 1c \}.$
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Tags: algebra, maximization, optimization, function, IMO Shortlist, IMO Longlist
vntbqpqh234
29.09.2010 15:18
i think that is easy. use $ \frac{\left | a \right |}{\left | b \right |}= \left | \frac{a}{b} \right |.$ and use $\frac{1}{a}$ is max
Pirkuliyev Rovsen
03.10.2010 18:59
Here my solution :We known $ \abs{x-y}+x+y=2max(x,y)$.Hence $f(a,b,c)=\abs{2max(\frac{1}{a},\frac{1}{b}-\frac{2}{c}}+2max(\frac{1}{a},\frac{1}{b})+\frac{2}{c} f(a,b,c)=2 max[2 max(\frac{1}{a},\frac{1}{b}=; \;\frac{2}{c} ]=4 max [max(\frac{1}{a};\frac{1}{b}) \; \frac{1}{c}]=4 max(\frac{1}{a};\frac{1}{b};\frac{1}{c}) $ Q.E.D
hrithikguy
19.09.2011 01:30
Note that $\max \{x,y\} = \dfrac {|x-y| +x+y}{2}$.
Thus, letting $x = \dfrac {|b-a|}{|ab|} + \dfrac {b+a}{ab}$, and $y = \dfrac {2}{c}$, we see that
$f(a,b,c) = 2\max \{\dfrac {|b-a|}{|ab|} + \dfrac {b+a}{ab}, \dfrac {2}{c}\}$.
Note that $\dfrac {|b-a|}{|ab|} + \dfrac {b+a}{ab} = \left |\dfrac {1}{a} - \dfrac {1}{b}\right | + \dfrac {1}{a} + \dfrac {1}{b}$.
Thus, this is equal to $2 \max \{\dfrac {1}{a}, \dfrac {1}{b} \}$.
Hence,
$f(a,b,c) = 2\max \{2\max \{\dfrac {1}{a}, \dfrac {1}{b} \}, \dfrac {2}{c} \}$.
Pulling the two out, we can easily
$f(a,b,c) =4\max \{\dfrac {1}{a}, \dfrac {1}{b}, \dfrac {1}{c} \}$, as desired.