Prove :1/a^{2n+1}+1/b^{2n+1}+1/c^{2n+1}=1/{a^{2n+1}+b^{2n+1}+c^{2n+1}}

[Albert1]

$given$
$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}$
$prove :$
$if \,\, n\in N\,\, then $
$\dfrac{1}{a^{2n+1}}+\dfrac{1}{b^{2n+1}}+\dfrac{1}{c^{2n+1}}=\dfrac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}$

 

[kaliprasad]

$given$
$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}$
$prove :$
$if \,\, n\in N\,\, then $
$\dfrac{1}{a^{2n+1}}+\dfrac{1}{b^{2n+1}}+\dfrac{1}{c^{2n+1}}=\dfrac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}$

Albert and others Happy new year

Spoiler

we have
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\dfrac{1}{a+b+c}$
$=>\frac{1}{a}+\frac{1}{b} = \frac{1}{a+b+c}- \frac{1}{c}$
$=>\frac{a+b}{ab} = - \frac{a+b}{c(a+b+c)}$
$=>(a+b) = 0\cdots(1)$ or $\frac{1}{ab} + \frac{1}{c(+b+c)} = 0$
$\frac{1}{ab} + \frac{1}{c(+b+c)} = 0$
$=>c(a+b+c) + ab = 0$
or $(a+c)(b+c) = 0$
so from (1) and above we have $a= -b$ or $b= -c$ or $c= - a$
beacuse of symmetry taking $a = -b$ we have a + b+ c = c and both LHS and RHS of given expression $\frac{1}{c^{2n+1}}$ and same
hence proved

 

[Albert1]

Albert and others Happy new year

Spoiler

we have
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\dfrac{1}{a+b+c}$
$=>\frac{1}{a}+\frac{1}{b} = \frac{1}{a+b+c}- \frac{1}{c}$
$=>\frac{a+b}{ab} = - \frac{a+b}{c(a+b+c)}$
$=>(a+b) = 0\cdots(1)$ or $\frac{1}{ab} + \frac{1}{c(+b+c)} = 0$
$\frac{1}{ab} + \frac{1}{c(+b+c)} = 0$
$=>c(a+b+c) + ab = 0$
or $(a+c)(b+c) = 0$
so from (1) and above we have $a= -b$ or $b= -c$ or $c= - a$
beacuse of symmetry taking $a = -b$ we have a + b+ c = c and both LHS and RHS of given expression $\frac{1}{c^{2n+1}}$ and same
hence proved

Happy new year to all MHB folks