Cho a,b,c >0 thỏa mãn : `a+1/b` =`b+1/c`=`c+1/a`
Chứng minh : `a^2005+1/b^2006`= `b^2005+1/c^2006`=`c^2005+1/a^2006`
Cho a,b,c >0 thỏa mãn : `a+1/b` =`b+1/c`=`c+1/a`
Chứng minh : `a^2005+1/b^2006`= `b^2005+1/c^2006`=`c^2005+1/a^2006`
`a+1/b=b+1/c=c+1/a`
\begin{cases}a+\frac{1}{b}=b+\frac{1}{c}\\b-\frac{1}{c}=c+\frac{1}{a}\\a+\frac{1}{b}=c+\frac{1}{a}\\\end{cases}
\begin{cases}a-b=\frac{(b-c)}{bc}\\b-c=\frac{(c-a)}{ac}\\a-c=\frac{b-a}{ab}\\\end{cases}
`⇒(a-b)(b-c)(a-c)=((a-b)(b-c)(a-c))/((abc))^2`
`⇒(a-b)(b-c)(a-c)(a^2b^2c^2-1)=0`
⇒\(\left[ \begin{array}{l}a=b=c\\abc=1\\abc=-1\end{array} \right.\)
`⇒ĐPCM`