Let $$f(a,b,c)=\left|\dfrac{|b-a|}{|ab|}+\dfrac{b+a}{ab}-2c\right|+\dfrac{|b-a|}{|ab|}+\dfrac{b+a}{ab}+\dfrac{2}{c}.$$ Find closed form to $f$.
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1And exactly why don't you consider this form 'closed'? – ajotatxe Feb 23 '16 at 16:54
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Could you tell which is the math contest this question originates, if such is the case? – Jean Marie Feb 23 '16 at 17:08
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I think it would be more appropriate to say simplify the following expression. – arbitUser1401 Feb 23 '16 at 22:24
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Note that $|x-y|+x+y=2\max\{x,y\}$ then $$f(a,b,c)=\left|\left|\dfrac{1}{a}-\dfrac{1}{b}\right|+\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{2}{c}\right|+\left|\dfrac{1}{a}-\dfrac{1}{b}\right|+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{c}=$$$$=|2\max\{1/a,1/b\}-\dfrac{2}{c}|+2\max\{1/a,1/b\}+\dfrac{2}{c}=$$$$=2\max[2\max\{1/a,1/b\};2/c]=4\max[\max\{1/a,1/b\};1/c]=4\max\{1/a,1/b,1/c\}.$$
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