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Question Number 206322 by MATHEMATICSAM last updated on 12/Apr/24
If abc = 1 then prove that   (1/(1 + a + b^(−1) )) + (1/(1 + b + c^(−1) )) + (1/(1 + c + a^(−1) )) = 1
$$\mathrm{If}\:{abc}\:=\:\mathrm{1}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{1}\:+\:{a}\:+\:{b}^{−\mathrm{1}} }\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:{b}\:+\:{c}^{−\mathrm{1}} }\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:{c}\:+\:{a}^{−\mathrm{1}} }\:=\:\mathrm{1} \\ $$
Commented by Rasheed.Sindhi last updated on 12/Apr/24
For similar question see  Q#205380
$${For}\:{similar}\:{question}\:{see} \\ $$$${Q}#\mathrm{205380} \\ $$
Answered by som(math1967) last updated on 12/Apr/24
(b/(b+ab+1))+(1/(1+b+ab))+((ab)/(ab+abc+b))  [ abc=1 ∴ab=c^(−1) ]  =(b/(b+1+ab))+(1/(b+1+ab)) +((ab)/(b+1+ab))  =((b+1+ab)/(b+1+ab))=1
$$\frac{{b}}{{b}+{ab}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{1}+{b}+{ab}}+\frac{{ab}}{{ab}+{abc}+{b}} \\ $$$$\left[\:{abc}=\mathrm{1}\:\therefore{ab}={c}^{−\mathrm{1}} \right] \\ $$$$=\frac{{b}}{{b}+\mathrm{1}+{ab}}+\frac{\mathrm{1}}{{b}+\mathrm{1}+{ab}}\:+\frac{{ab}}{{b}+\mathrm{1}+{ab}} \\ $$$$=\frac{{b}+\mathrm{1}+{ab}}{{b}+\mathrm{1}+{ab}}=\mathrm{1} \\ $$

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