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

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The discussion centers on proving the equation involving the reciprocals of powers of three variables a, b, and c. It states that if the initial condition holds true, then the relationship for higher odd powers also holds for natural numbers n. Participants express greetings and share New Year wishes, indicating a friendly atmosphere. The mathematical proof is the primary focus, emphasizing the connection between the two equations. The thread highlights the importance of establishing this relationship in mathematical contexts.
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$given$
$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}$
$prove :$
$if \,\, n\in N\,\, then $
$\dfrac{1}{a^{2n+1}}+\dfrac{1}{b^{2n+1}}+\dfrac{1}{c^{2n+1}}=\dfrac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}$
 
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Albert said:
$given$
$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}$
$prove :$
$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

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
 
kaliprasad said:
Albert and others Happy new year

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
 
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