Ta có:\(a^2+b^2+c^2=2\)

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

                   Mà a+b+c=2

                        \(\Rightarrow4-2ab-2ac-2bc=2\)

                         \(\Rightarrow2-2ab-2ac-2bc=0\)

                         \(\Rightarrow-2\left(ab+ac+bc\right)=-2\)

                         \(\Rightarrow ab+ac+bc=1\left(1\right)\)

Ta lại có:\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ab+ac+bc}{abc}\)

                      Từ (1) suy ra đc:\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\left(đpcm\right)\)

theo bài ra ta có: a+b+c=2 => (a+b+c)^2 =4 => a^2 +b^2 +c^2 +2(ab+bc+ca)=4=> 2(ab+bc+ca)=2(vì a^2 +b^2 +c^2=2) 

=> ab+bc+ca=1   =>\(\frac{ab}{abc}+\frac{bc}{abc}+\frac{ca}{abc}=\frac{1}{abc}\)        (vì abc khác 0)

                          => \(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}=\frac{1}{abc}\)

Vậy với a+b+c=a^2+b^2+c^2=2 và abc khác  0 thì \(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}=\frac{1}{abc}\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\) (ĐKXĐ: \(a\ne0;b\ne0;c\ne0;a+b+c\ne0\))

<=> \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{a+b+c}=0\)

<=> \(\dfrac{b}{ab}+\dfrac{a}{ab}+\dfrac{a+b+c}{c\left(a+b+c\right)}-\dfrac{c}{c\left(a+b+c\right)}=0\)

<=> \(\dfrac{a+b}{ab}+\dfrac{a+b}{c\left(a+b+c\right)}=0\)

<=> \(\left(a+b\right)\left(\dfrac{1}{ab}+\dfrac{1}{c\left(a+b+c\right)}\right)=0\)

<=> \(\left(a+b\right)\left[\dfrac{c\left(a+b+c\right)}{abc\left(a+b+c\right)}+\dfrac{ab}{abc\left(a+b+c\right)}\right]=0\)

<=> \(\dfrac{\left(a+b\right)\left(b+c\right)\left(a+c\right)}{abc\left(a+b+c\right)}=0\) [vì c(a + b + c) + ab = ac + bc + c² + ab = a(b + c) + c(b + c) = (a + c)(b + c)]

<=> (a + b)(b + c)(a + c) = 0

câu c : vì nhân hai vế ta được :

(a+b+c)x (ab+bc+ac)=abc

abc+a\(^2\)b+\(a^2\)c + b^2c+ab^2+abc+bc^2+ac^c+abc=abc

abc+a^2b+a^2c+ b^2c+ab^2+abc+bc^2+ac^c=0

(a+c)(a+b)(b+c)=0

Áp dụng BDT svacxơ ta có:

 \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}=\frac{9}{1}=9\)

Vì \(a+b+c=1\)

Dấu ''='' khi a=b=c

Học tốt.

bạn làm theo cách bđt cosi giúp mình được không ạ?

Dề sai thế \(a=\frac{1}{3};b=5;c=\frac{3}{5}\)vô đi nhé.

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

\(\frac{a+b}{c}-1=\frac{b+c}{a}-1=\frac{c+a}{b}-1\)

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

ADTCDTSBN,ta có

\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{c+a+b}=2\) 

P=(1+b/a)³

Cm b/a=c/d=a/c