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a/(1+a) +b/(1+b) +c/(1+c)=1a,b,c are positive real nosP.T.abcabc is less than equal to 1/8

Aryan Singh , 8 Years ago
Grade 10
anser 2 Answers
jagdish singh singh

Last Activity: 8 Years ago

$\hspace{-0.7 cm}Let $\frac{a}{1+a}=x\Rightarrow a = \frac{x}{1-x}$ Similarly $\frac{b}{1+b} = y\Rightarrow b=\frac{y}{1-y}$ \\\\ Similarly $\frac{c}{1+c}=z\Rightarrow c=\frac{z}{1-z}$ and given $a,b,c>0$\\\\ So $0<x,y,z<1\;,$ Now we have prove $\frac{x}{1-x}\cdot \frac{y}{1-y}\cdot \frac{z}{1-z}\leq \frac{1}{8}$\\\\ So proving $(1-x)(1-y)(1-z)\geq 8xyz$ given $x+y+z=1$\\\\ So proving $(y+z)(z+x)(x+y)\geq 8xyz$\\\\ Using $\bf{A.M\geq G.M}\;,$ We get $(x+y)(y+z)(z+x)\geq 8xyz.$

Yogesh Kumar

Last Activity: 7 Years ago

  • a/(1+a)+b/(1+b)+c/(1+c)=1
 
on simplifying it we get,
 
  • ab+bc+ca+2abc=1
 
Now Using AM ≥ GM,
  • (ab+bc+ca+2abc)÷4≥((2)(abc)3)¼
  • (1/4)4≥2(abc)3
  • (1/512)1/3≥abc or abc≤1/8
Hence Proved
 
 

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