Khánh Linh ơi , ghép đôi như thế này á : \(\frac{a^2}{b^2}+\frac{b^2}{c^2}\ge2\frac{a}{c}\)

                                                                  \(\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge2\frac{b}{a}\)

                                                                  \(\frac{c^2}{a^2}+\frac{a^2}{b^2}\ge2\frac{c}{b}\)

xong cộng vào theo vế đúng ko , nhưng mk thấy người ta vt đây là số thực ?

\(\frac{a}{b-c}=-\frac{b}{c-a}-\frac{c}{a-b}=-\frac{b\left(a-b\right)+c\left(c-a\right)}{\left(c-a\right)\left(a-b\right)}\Rightarrow\frac{a}{\left(b-c\right)^2}=-\frac{b\left(a-b\right)+c\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-c\right)}\)

sau đó chứng minh tương tự và cộng theo từng vế thôi 

Ta có

\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)

\(\Rightarrow\left\{\begin{matrix}\frac{a}{b-c}=-\frac{b}{c-a}-\frac{c}{a-b}\\\frac{b}{c-a}=-\frac{a}{b-c}-\frac{c}{a-b}\\\frac{c}{a-b}=-\frac{a}{b-c}-\frac{b}{c-a}\end{matrix}\right.\) (1)

Mà

\(\left\{\begin{matrix}\frac{a}{\left(b-c\right)^2}=\frac{a}{b-c}.\frac{1}{b-c}\\\frac{b}{\left(c-a\right)^2}=\frac{b}{c-a}.\frac{1}{c-a}\\\frac{c}{\left(a-b\right)^2}=\frac{c}{a-b}.\frac{1}{a-b}\end{matrix}\right.\)

Ta có : \(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)

\(\Rightarrow\frac{a}{b-c}.\frac{1}{b-c}+\frac{b}{c-a}.\frac{1}{c-a}+\frac{c}{a-b}.\frac{1}{a-b}=0\)

Thay điều (1) vào biểu thức ta có :

\(\frac{a}{b-c}.\frac{1}{b-c}+\frac{b}{c-a}.\frac{1}{c-a}+\frac{c}{a-b}.\frac{1}{a-b}=0\)

\(\Rightarrow\left(-\frac{b}{c-a}-\frac{c}{a-b}\right).\frac{1}{b-c}+\left(-\frac{a}{b-c}-\frac{c}{a-b}\right).\frac{1}{c-a}+\left(-\frac{a}{b-c}-\frac{b}{c-a}\right).\frac{1}{a-b}=0\)

\(\Rightarrow-\frac{b}{\left(c-a\right)\left(b-c\right)}-\frac{c}{\left(a-b\right)\left(b-c\right)}-\frac{a}{\left(b-c\right)\left(c-a\right)}-\frac{c}{\left(a-b\right)\left(c-a\right)}-\frac{a}{\left(b-c\right)\left(a-b\right)}-\frac{b}{\left(c-a\right)\left(a-b\right)}=0\)

\(\Rightarrow-\frac{b}{\left(c-a\right)\left(b-c\right)}-\frac{a}{\left(c-a\right)\left(b-c\right)}-\frac{c}{\left(a-b\right)\left(b-c\right)}-\frac{a}{\left(a-b\right)\left(b-c\right)}-\frac{c}{\left(c-a\right)\left(a-b\right)}-\frac{b}{\left(c-a\right)\left(a-b\right)}=0\)

\(\Rightarrow-\frac{b-a}{\left(c-a\right)\left(b-c\right)}-\frac{c-a}{\left(a-b\right)\left(b-c\right)}-\frac{c-b}{\left(c-a\right)\left(a-b\right)}=0\)

\(\Rightarrow-\left[\frac{b+a}{\left(c-a\right)\left(b-c\right)}+\frac{c+a}{\left(a-b\right)\left(b-c\right)}+\frac{c+b}{\left(c-a\right)\left(a-b\right)}\right]=0\)

\(\Rightarrow-\left[\frac{\left(b+a\right)\left(a-b\right)^2\left(b-c\right)\left(c-a\right)+\left(c+a\right)\left(c-a\right)^2\left(b-c\right)\left(a-b\right)+\left(c+b\right)\left(b-c\right)^2\left(c-a\right)\left(a-b\right)}{\left(b-c\right)^2\left(c-a\right)^2\left(a-b\right)^2}\right]=0\)

\(\Rightarrow-\left\{\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left[\left(b+a\right)\left(a-b\right)+\left(c+a\right)\left(c-a\right)+\left(b+c\right)\left(b-c\right)\right]}{\left(b-c\right)^2\left(c-a\right)^2\left(a-b\right)^2}\right\}=0\)

\(\Rightarrow-\left[\frac{\left(b+a\right)\left(b-a\right)+\left(c+a\right)\left(c-a\right)+\left(b+c\right)\left(b-c\right)}{\left(b-c\right)\left(c-a\right)\left(a-b\right)}\right]=0\)

\(\Rightarrow-\left[\frac{\left(a^2-b^2\right)+\left(c^2-a^2\right)+\left(b^2-c^2\right)}{\left(b-c\right)\left(c-a\right)\left(a-b\right)}\right]=0\)

\(\Rightarrow-\left[\frac{\left(-b^2+b^2\right)+\left(-a^2+a^2\right)+\left(-c^2+c^2\right)}{\left(b-c\right)\left(c-a\right)\left(a-b\right)}\right]=0\)

\(\Rightarrow-\left[\frac{0}{\left(b-c\right)\left(c-a\right)\left(a-b\right)}\right]=0\)

\(\Rightarrow0=0\) ( đpcm )

Đặt \(\left(\frac{a-b}{c},\frac{b-c}{a},\frac{c-a}{b}\right)\rightarrow\left(x,y,z\right)\)

Khi đó:\(\left(\frac{c}{a-b},\frac{a}{b-c},\frac{b}{c-a}\right)\rightarrow\left(\frac{1}{x},\frac{1}{y},\frac{1}{z}\right)\)

Ta có:

\(P\cdot Q=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)

Mặt khác:\(\frac{y+z}{x}=\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\cdot\frac{c}{a-b}=\frac{b^2-bc+ac-a^2}{ab}\cdot\frac{c}{a-b}\)

\(=\frac{c\left(a-b\right)\left(c-a-b\right)}{ab\left(a-b\right)}=\frac{c\left(c-a-b\right)}{ab}=\frac{2c^2}{ab}\left(1\right)\)

Tương tự:\(\frac{x+z}{y}=\frac{2a^2}{bc}\left(2\right)\)

\(=\frac{x+y}{z}=\frac{2b^2}{ac}\left(3\right)\)

Từ ( 1 );( 2 );( 3 ) ta có:
\(P\cdot Q=3+\frac{2c^2}{ab}+\frac{2a^2}{bc}+\frac{2b^2}{ac}=3+\frac{2}{abc}\left(a^3+b^3+c^3\right)\)

Ta có:\(a+b+c=0\)

\(\Rightarrow\left(a+b\right)^3=-c^3\)

\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)

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

Khi đó:\(P\cdot Q=3+\frac{2}{abc}\cdot3abc=9\)

Mách mk nốt 2 bài kia vs

chiju

Em thử nha, có gì sai bỏ qua ạ.

Đề cho gọn,Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\) thì \(xy+yz+zx=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{a+b+c}{abc}=0\) 

Và \(x+y+z=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ab+bc+ca}{abc}=0\)

Ta có: \(VT=\sqrt{x^2+y^2+z^2}=\sqrt{\left(x+y+z\right)^2-2\left(xy+yz+zx\right)}=0\) (1)

Mặt khác,ta có \(VT=\left|x+y+z\right|=0\) (2)

Từ (1) và (2) ta có đpcm

​Dòng cuối phải là

VP=|x+y+z|=0 

đúng không????

uk,thì e viết nhầm -_-"

Xét \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}\)

\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)}\)

\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2\left(\frac{a+b+c}{abc}\right)}\)

\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}=|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}|\)

\(\Rightarrow\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}|\)(đpcm)