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

Ta có:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) = \(\overline{\frac{\overline{bc}+\overline{ac}+\overline{ac}}{\overline{abc}}}\) = ab + bc + ca 
=> a + b + c = ab + bc + ca 
=> a + b + c - ab - bc - ca = 0 
=> a + b + c - ab - bc - ac + abc - 1 = 0 
=> (a - ab) + (b - 1) + (c - bc) + (abc - ac) = 0 
=> - a(b - 1) + (b - 1) - c(b - 1) + ac(b - 1) = 0 
=> (b - 1)(- a + 1 - c + ac) = 0 
=> (b - 1)[( - a + 1) + (ac - c)] = 0 
=> (b - 1)[ - (a - 1) + c(a - 1)] = 0 
=> (a - 1)(b - 1)(c - 1) = 0 
=> a - 1 = 0 hoặc b - 1 = 0 hoặc c - 1 = 0 
=> a = 1 hoặc b = 1 hoặc c = 1 

Vậy (a - 1)(b - 1)(c - 1) > 1

\(\left(a-1\right)\left(b-1\right)\left(c-1\right)>0\)

\(\Leftrightarrow\left(ab-a-b+1\right)\left(c-1\right)>0\)

\(\Leftrightarrow abc-ac-bc+c-ab+a+b-1>0\)

\(\Leftrightarrow-ab-bc-ab+a+b+c>0\)

\(\Leftrightarrow a+b+c>ab+ac+bc\)

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

\(\Leftrightarrow a+b+c>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) (thỏa mãn đề bài)

Vậy \(\left(a-1\right)\left(b-1\right)\left(c-1\right)>0\)

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

\(\dfrac{a}{a'}+\dfrac{b'}{b}=1\Rightarrow\dfrac{a}{a'}\cdot\dfrac{b}{b'}+\dfrac{b'}{b}\cdot\dfrac{b}{b'}=\dfrac{b}{b'}\Rightarrow\dfrac{ab}{a'b'}+1=\dfrac{b}{b'}\left(1\right)\)

\(\dfrac{b}{b'}+\dfrac{c'}{c}=1\Rightarrow\dfrac{b}{b'}=1-\dfrac{c'}{c}\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow\dfrac{ab}{a'b'}=-\dfrac{c'}{c}\Rightarrow abc=-a'b'c'\Rightarrow abc+a'b'c'=0\)

Vậy \(abc+a'b'c'=0\left(dpcm\right)\)

Help me vs mấy chế ơi

Lớp 9 chưa học cauchy thì làm cách này nha :v

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}\)

\(=1+\dfrac{b}{a}+\dfrac{c}{a}+1+\dfrac{a}{b}+\dfrac{c}{b}+1+\dfrac{a}{c}+\dfrac{b}{c}\)

\(=3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\ge3+2+2+2=9\)

\(-->đpcm\) \("="\) khi \(a=b=c=\dfrac{1}{3}\)

áp dụng cauchy-schwarz dạng engel ta có :

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}=9\left(đpcm\right)\)

Ngược dấu rồi

Mk sửa r đó. H giúp mk vs. Cảm ơn

Cô-si mẫu suy ra: 

\(A\le\frac{1}{2}\left(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\right)\)

Dễ cm biểu thức trong ngoặc = 1. 

Suy ra A <=1/2

Dấu = khi a=b=c=1