Lời giải:
Ta có:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)
\(\Leftrightarrow \frac{a+b}{ab}+\frac{a+b}{c(a+b+c)}=0\)
\(\Leftrightarrow (a+b)\left(\frac{1}{ab}+\frac{1}{c(a+b+c)}\right)=0\)
\(\Leftrightarrow \frac{(a+b)[c(a+b+c)+ab]}{abc(a+b+c)}=0\)
\(\Leftrightarrow (a+b)(b+c)(c+a)=0\)
Xét : \(A=\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}-\frac{1}{a^n+b^n+c^n}\)
\(A=\frac{a^n+b^n}{a^nb^n}+\frac{a^n+b^n}{c^n(a^n+b^n+c^n)}\)
\(A=(a^n+b^n)\left(\frac{1}{a^nb^n}+\frac{1}{c^n(a^n+b^n+c^n)}\right)\)
\(A=\frac{(a^n+b^n)[c^n(a^n+b^n+c^n)+a^nb^n]}{a^nb^nc^n(a^n+b^n+c^n)}\)
\(A=\frac{(a^n+b^n)(b^n+c^n)(c^n+a^n)}{a^nb^nc^n(a^n+b^n+c^n)}\)
Vì $n$ lẻ nên :
\((a^n+b^n)(b^n+c^n)(c^n+a^n)=(a+b)(b+c)(c+a)(a^{n-1}+....+b^{n-1})(b^{n-1}+..+c^{n-1})(c^{n-1}+...+a^{n-1})\)
\(=0\) do \((a+b)(b+c)(c+a)=0\)
Do đó: \(A=0\Leftrightarrow \frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{1}{a^n+b^n+c^n}\)
