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Question Number 36978 by jayanta11 last updated on 07/Jun/18
if (a^2 /(b+c)) = (b^2 /(c+a)) = (c^2 /(a+b)) = 1 then find the value of (1/(1+a)) + (1/(1+b)) + (1/(1+c))
$${if}\:\:\frac{{a}^{\mathrm{2}} }{{b}+{c}}\:=\:\frac{{b}^{\mathrm{2}} }{{c}+{a}}\:=\:\frac{{c}^{\mathrm{2}} }{{a}+{b}}\:=\:\mathrm{1}\:{then}\:{find}\:{the}\:{value}\:{of}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{a}}\:+\:\frac{\mathrm{1}}{\mathrm{1}+{b}}\:+\:\frac{\mathrm{1}}{\mathrm{1}+{c}} \\ $$
Commented by behi83417@gmail.com last updated on 07/Jun/18
a^2 =b+c,b^2 =c+a,c^2 =a+b ⇒a^2 +b^2 +c^2 =2a+2b+2c (a−1)^2 +(b−1)^2 +(c−1)^2 =3⇒ a−1=b−1=c−1=1⇒a=b=c=2 Σ(1/(1+a))=3×(1/3)=1 or: (1/(1+a))=((a^2 /(b+c))/(a+(a^2 /(b+c))))=(a^2 /(a^2 +a(b+c)))=(a/(a+b+c)) ⇒Σ(1/(1+a))=Σ(a/(a+b+c))=((a+b+c)/(a+b+c))=1 .■
$${a}^{\mathrm{2}} ={b}+{c},{b}^{\mathrm{2}} ={c}+{a},{c}^{\mathrm{2}} ={a}+{b} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{2}{a}+\mathrm{2}{b}+\mathrm{2}{c} \\ $$$$\left({a}−\mathrm{1}\right)^{\mathrm{2}} +\left({b}−\mathrm{1}\right)^{\mathrm{2}} +\left({c}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{3}\Rightarrow \\ $$$${a}−\mathrm{1}={b}−\mathrm{1}={c}−\mathrm{1}=\mathrm{1}\Rightarrow{a}={b}={c}=\mathrm{2} \\ $$$$\Sigma\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{a}}}=\mathrm{3}×\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{1} \\ $$$$\boldsymbol{{or}}: \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{a}}=\frac{\frac{{a}^{\mathrm{2}} }{{b}+{c}}}{{a}+\frac{{a}^{\mathrm{2}} }{{b}+{c}}}=\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{a}\left({b}+{c}\right)}=\frac{{a}}{{a}+{b}+{c}} \\ $$$$\Rightarrow\Sigma\frac{\mathrm{1}}{\mathrm{1}+{a}}=\Sigma\frac{{a}}{{a}+{b}+{c}}=\frac{{a}+{b}+{c}}{{a}+{b}+{c}}=\mathrm{1}\:.\blacksquare \\ $$
Answered by ajfour last updated on 07/Jun/18
a^2 +a=b^2 +b=c^2 +c=a+b+c=s ⇒a(a+1)=b(b+1)=c(c+1)=s (1/(1+a))+(1/(1+b))+(1/(1+c))=(a/s)+(b/s)+(c/s)=1 .
$${a}^{\mathrm{2}} +{a}={b}^{\mathrm{2}} +{b}={c}^{\mathrm{2}} +{c}={a}+{b}+{c}=\boldsymbol{{s}} \\ $$$$\Rightarrow{a}\left({a}+\mathrm{1}\right)={b}\left({b}+\mathrm{1}\right)={c}\left({c}+\mathrm{1}\right)=\boldsymbol{{s}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{a}}}+\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{b}}}+\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{c}}}=\frac{\boldsymbol{{a}}}{\boldsymbol{{s}}}+\frac{\boldsymbol{{b}}}{\boldsymbol{{s}}}+\frac{\boldsymbol{{c}}}{\boldsymbol{{s}}}=\mathrm{1}\:. \\ $$

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