If $$ abc=1 $$ , prove $$ \dfrac{a}{1+a+ab}+\dfrac{b}{1+b+bc}+\dfrac{c}{1+c+ca} $$ =1

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If $$ abc=1 $$ , prove $$ \dfrac{a}{1+a+ab}+\dfrac{b}{1+b+bc}+\dfrac{c}{1+c+ca} $$  =1

Uniteasy Admin · Accepted Answer

$$\because abc = 1$$,
$$ \dfrac{a}{1+a+ab} =\dfrac{a}{abc+a+ab} = \dfrac{1}{1+b+bc}  $$
 $$	herefore \dfrac{a}{1+a+ab}+\dfrac{b}{1+b+bc}+\dfrac{c}{1+c+ca} $$
$$= \dfrac{1}{1+b+bc}+\dfrac{b}{1+b+bc}+\dfrac{c}{1+c+ca} $$
$$=\dfrac{1+b}{1+b+bc} +\dfrac{c}{1+c+ca}$$
$$=\dfrac{1+b+bc-bc}{1+b+bc} +\dfrac{c}{1+c+ca}$$
$$=1-\dfrac{bc}{1+b+bc} +\dfrac{c}{1+c+ca}$$
$$=1-\dfrac{bc}{abc+b+bc} +\dfrac{c}{1+c+ca}$$
$$=1$$

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