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Question Number 205380 by cortano12 last updated on 19/Mar/24

Answered by A5T last updated on 19/Mar/24

$$\frac{\mathrm{\Sigma }(1+bc+b)(1+ca+c)=\mathrm{\Sigma }(2+ca+2c+2bc+{bc}^2+b)}{(2+ca+2c+2bc+{bc}^2+b)(1+ab+a)}$$$$=\frac{6+3(ca+ab+bc)+3(a+b+c)+{bc}^2+{ca}^2+{ab}^2}{6+3ca+3c+3bc+{bc}^2+3b+3ab+3a+a^{2}c+{ab}^2}$$$$=1$$

Answered by Rasheed.Sindhi last updated on 19/Mar/24

$$abc=1$$$$\frac{1}{1+ab+a}+\frac{1}{1+bc+b}+\frac{1}{1+ca+c}$$$$=\frac{1\left(c\right)}{(1+ab+a)\left(c\right)}+\frac{1\left(ca\right)}{(1+bc+b)\left(ca\right)}+\frac{1\times 1}{(1+ca+c)\left(1\right)}$$$$=\frac{c}{c+abc+ca}+\frac{ca}{ca+abc.c+abc}+\frac{1}{1+ca+c}$$$$=\frac{c}{c+1+ca}+\frac{ca}{ca+c+1}+\frac{1}{1+ca+c}$$$$=\frac{c+ca+1}{c+ca+1}=1$$$$$$

Answered by Rasheed.Sindhi last updated on 19/Mar/24

$$abc=1\Rightarrow \{\begin{array}{l}bc=\frac{1}{a}\\ c=\frac{1}{ab}\end{array}$$$$\frac{1}{1+ab+a}+\frac{1}{1+bc+b}+\frac{1}{1+ca+c}$$$$=\frac{1}{1+ab+a}+\frac{1}{1+\frac{1}{a}+b}+\frac{1}{1+ca+\frac{1}{ab}}$$$$\frac{1}{1+ab+a}+\frac{a}{a+1+ab}+\frac{ab}{ab+a.abc+1}$$$$=\frac{1+a+ab}{1+a+ab}=1$$

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