2. Áp dụng bất đẳng thức Cô - si cho 3 số dương \(\frac{a}{b},\frac{b}{c},\frac{c}{a}\)ta có

\(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3\sqrt[3]{\frac{a}{b}.\frac{b}{c}.\frac{c}{a}}\)\(=3\)

Dấu "=" xảy ra <=> a = b = c

mk chỉnh lại đề nhé: 

cho \(a^3+b^3+c^3=3abc\)

CMR: \(a=b=c\)hoặc   \(a+b+c=0\)

      BÀI LÀM

\(a^3+b^3+c^3=3abc\)

<=>  \(a^3+b^3+c^3-3abc=0\)

<=> \(\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)

<=> \(\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)

<=> \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

<=>  \(\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)

Xét:  \(a^2+b^2+c^2-ab-bc-ca=0\)

<=>  \(2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

<=>  \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

<=>  \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\)

<=>  \(a=b=c\)

=> đpcm

Ta có

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

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

Áp dụng cô si cho từng cặp

\(\frac{a}{c}+\frac{c}{a}\ge2;\frac{a}{b}+\frac{b}{a}\ge2;\frac{b}{c}+\frac{c}{b}\ge2\)

=>....

Dấu = xảy ra <=>a=b=c

8)

a)6-|x|=2

=>6-x=2

x=6-2

x=4

b)6+|x|=2

=>6+x=2

x=2-6

x=-4

Chịu !!