Giải thích các bước giải:

Ta có:

$\dfrac1a+\dfrac1b+\dfrac1c=\dfrac1{a+b+c}$

$\to \dfrac1a+\dfrac1b+\dfrac1c-\dfrac1{a+b+c}=0$

$\to \dfrac{a+b}{ab}+\dfrac{a+b+c-c}{c(a+b+c)}=0$

$\to \dfrac{a+b}{ab}+\dfrac{a+b}{c(a+b+c)}=0$

$\to (a+b)\cdot (\dfrac1{ab}+\dfrac1{c(a+b+c)})=0$

$\to (a+b)\cdot \dfrac{ab+c(a+b+c)}{abc(a+b+c)}=0$

$\to (a+b)\cdot \dfrac{(c+a)(c+b)}{abc(a+b+c)}=0$

 $\to \dfrac{(a+b)(b+c)(c+a)}{abc(a+b+c)}=0$

$\to (a+b)(b+c)(c+a)=0$

$\to a+b=0$ hoặc $b+c=0$ hoặc $c+a=0$

Giải $a+b=0$

$\to a=-b$

$\to \dfrac1{a^{2013}}+\dfrac1{b^{2013}}+\dfrac1{c^{2013}}=\dfrac1{a^{2013}}-\dfrac1{a^{2013}}+\dfrac1{c^{2013}}=\dfrac1{c^{2013}}=\dfrac1{a^{2013}+b^{2013}+c^{2013}}$

Tương tự hai trường hợp còn lại

$\to đpcm$