Bài làm

8x² - 12xy - 8y² = 0

=> ( 8x² - 8y² ) - 12xy = 0

=> 8( x² - y² ) - 12xy = 0

=> 8( x - y )( x + y ) - 12xy = 0

=> 4[ 2( x - y )( x + y ) - 3xy ] = 0

Ta có BĐT \(\frac{x}{y}+\frac{y}{x}\ge2\forall x,y\ge0\)

\(\Rightarrow A=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge2+2+2=6\)

Dấu "=" xảy ra khi \(a=b=c\)

cái này dùng co si xong luôn nha bạn

\(A=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\)

\(A=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\)

vì a;b;c dương nên \(\frac{a}{b};\frac{b}{a};\frac{b}{c};\frac{c}{b};\frac{a}{c};\frac{c}{a}>0\)

áp dụng bđt cô si ta có : 

\(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{a}{b}\cdot\frac{b}{a}}=2\)

\(\frac{b}{c}+\frac{c}{b}\ge2\sqrt{\frac{b}{c}\cdot\frac{c}{b}}=2\)

\(\frac{a}{c}+\frac{c}{a}\ge2\sqrt{\frac{a}{c}\cdot\frac{c}{a}}=2\)

\(\Rightarrow A\ge8\)

dấu = xảy ra khi \(\hept{\begin{cases}\frac{a}{b}=\frac{b}{a}\\\frac{b}{c}=\frac{c}{b}\\\frac{a}{c}=\frac{c}{a}\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\a=c\end{cases}\Leftrightarrow a=b=c}\)

1+1=2 UwU