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- #1

- Feb 14, 2012

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- Thread starter anemone
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- Thread starter
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- #1

- Feb 14, 2012

- 3,681

- Jan 25, 2013

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$\dfrac {c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}=1----(1)$

let :

$\dfrac {c^2}{a+b}+\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}=k$

$(1)\times a+(1)\times b+(1)\times c$

we get

k+a+b+c=a+b+c

$\therefore k=0$

let :

$\dfrac {c^2}{a+b}+\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}=k$

$(1)\times a+(1)\times b+(1)\times c$

we get

k+a+b+c=a+b+c

$\therefore k=0$

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