Question

[EnderCraft Gaming]

Cho \(a,b,c>0\)thoả mãn \(a.\left(\frac{1}{b}+\frac{1}{c}\right)=b.\left(\frac{1}{a}+\frac{1}{c}\right)=c.\left(\frac{1}{a}+\frac{1}{b}\right)\)Chứng minh rằng \(a=b=c\)

Answer 1

[Trí Tiên]

P/s: Bài toán này khá hay đó !!

Ta có : \(a\left(\frac{1}{b}+\frac{1}{c}\right)=b\left(\frac{1}{a}+\frac{1}{c}\right)=c\left(\frac{1}{a}+\frac{1}{b}\right)\)

\(\Leftrightarrow\frac{a^2c+a^2b}{abc}=\frac{b^2c+ab^2}{abc}=\frac{c^2b+c^2a}{abc}\)

Mà : \(a,b,c>0\)

\(\Rightarrow a^2c+a^2b=b^2c+ab^2=c^2b+c^2a\)

+) Xét : \(a^2c+a^2b=b^2c+ab^2\)

\(\Leftrightarrow ab\left(a-b\right)+c\left(a^2-b^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(ab+ca+cb\right)=0\)

\(\Leftrightarrow a-b=0\Leftrightarrow a=b\) (1)

( Do \(a,b,c>0\Rightarrow ab+ca+cb>0\) )

+) Xét \(b^2c+ab^2=c^2b+c^2a\)

\(\Leftrightarrow bc\left(b-c\right)+a\left(b^2-c^2\right)=0\)

\(\Leftrightarrow\left(b-c\right)\left(bc+ab+ac\right)=0\)

\(\Leftrightarrow b-c=0\Leftrightarrow b=c\)(2)

( Do \(a,b,c>0\Rightarrow ab+ca+cb>0\) )

Từ (1) và (2) \(\Rightarrow a=b=c\) (đpcm)

Answer 2

[EnderCraft Gaming]

 Thx nha !