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If $a+b+c=m$, prove that: $$\frac {x^{2a}}{x^{2a}+x^{m-b}+x^{m-c}} + \frac {x^{2b}}{x^{2b}+x^{m-c}+x^{m-a}}+ \frac {x^{2c}}{x^{2a}+x^{m-a}+x^{m-b}}=1.$$

My Attempt:

$a+b+c=m$

$a=m-b-c$

$b=m-a-c$

$c=m-a-b$,

L.H.S.$=\frac {x^{2a}}{x^{2a}+x^{a+c}+x^{a+b}}+\frac {x^{2b}}{x^{2b}+x^{a+b}+x^{b+c}}+\frac {x^{2c}}{x^{2c}+x^{b+c}+x^{a+c}}$.

Now, how should I move on?

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Note that

$$\frac{x^{2a}}{x^{2a}+x^{a+c}+x^{a+b}} = \frac{x^a}{x^a} \frac{x^a}{x^a+x^b+x^c}.$$

Using a similar reasoning for the other two terms, we find that

$$\begin{align*} &\frac {x^{2a}}{x^{2a}+x^{a+c}+x^{a+b}}+\frac {x^{2b}}{x^{2b}+x^{a+b}+x^{b+c}}+\frac {x^{2c}}{x^{2c}+x^{b+c}+x^{a+c}}\\ &= \frac{x^a}{x^a+x^b+x^c} + \frac{x^b}{x^a+x^b+x^c} + \frac{x^c}{x^a+x^b+x^c} \\ &= 1. \end{align*}$$

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Hint:

$$ \frac {x^{2a}}{x^{2a}+x^{m-b}+x^{m-c}} = \frac {x^{a}}{x^{a}+x^{m-b-a}+x^{m-c-a}} = \frac {x^{a}}{x^{a}+x^{b}+x^{c}} $$

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