Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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 .

$${a}^{\mathrm{2}} +{a}={b}^{\mathrm{2}} +{b}={c}^{\mathrm{2}} +{c}={a}+{b}+{c}=\boldsymbol{{s}} \\ $$$$\Rightarrow{a}\left({a}+\mathrm{1}\right)={b}\left({b}+\mathrm{1}\right)={c}\left({c}+\mathrm{1}\right)=\boldsymbol{{s}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{a}}}+\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{b}}}+\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{c}}}=\frac{\boldsymbol{{a}}}{\boldsymbol{{s}}}+\frac{\boldsymbol{{b}}}{\boldsymbol{{s}}}+\frac{\boldsymbol{{c}}}{\boldsymbol{{s}}}=\mathrm{1}\:. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com