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

=> bc+ac+ab=0

ta có

\(bc+ac=-ab\)

<=> \(\left(bc+ac\right)^2=a^2b^2\)

<=> \(b^2c^2+a^2c^2+2abc^2=a^2b^2\)

<=> \(b^2c^2+a^2c^2-a^2b^2=-2abc^2\)

tương tự

\(a^2b^2+b^2c^2-c^2a^2=-2ab^2c\)

\(c^2a^2+a^2b^2-b^2c^2=-2a^2bc\)

thay vào E ta đc

\(E=\dfrac{-a^2b^2c^2}{2ab^2c}-\dfrac{a^2b^2c^2}{2abc^2}-\dfrac{a^2b^2c^2}{2a^2bc}\)

=\(-\dfrac{ac}{2}-\dfrac{ab}{2}-\dfrac{bc}{2}=\dfrac{-\left(ac+ab+bc\right)}{2}=0\) (vì ac+bc+ab=0 cmt)

\(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=0\Leftrightarrow\frac{bc+ac-ab}{abc}=0\)

Vì \(a,b,c\ne0\Rightarrow abc\ne0\)

\(\Rightarrow bc+ac-ab=0\)

\(\Rightarrow\hept{\begin{cases}\left(bc+ac\right)^2=\left(ab\right)^2\\\left(bc-ab\right)^2=\left(-ac\right)^2\\\left(ac-ab\right)^2=\left(-bc\right)^2\end{cases}\Rightarrow\hept{\begin{cases}b^2c^2+c^2a^2-a^2b^2=-2abc^2\\b^2c^2+a^2b^2-a^2c^2=2ab^2c\\a^2c^2+a^2b^2-b^2c^2=2a^2bc\end{cases}}}\)

\(\Rightarrow E=\frac{a^2b^2c^2}{2ab^2c}+\frac{a^2b^2c^2}{-2abc^2}+\frac{a^2b^2c^2}{2a^2bc}\)

\(\Rightarrow E=\frac{ac}{2}-\frac{ab}{2}+\frac{bc}{2}=\frac{ac-ab+bc}{2}=\frac{0}{2}=0\)

CHÚC BẠN HỌC TỐT

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{bc+ac-ab}{abc}=0\)

Vì \(a,b,c\ne0\Rightarrow a.b.c\ne0\)

\(\Rightarrow bc+ac-ab=0\)

\(\Rightarrow\hept{\begin{cases}\left(bc+ac\right)^2=\left(ab\right)^2\\\left(bc-ab\right)^2=\left(-ac\right)^2\\\left(ac-ab\right)^2=\left(-bc\right)^2\end{cases}\Rightarrow}\hept{\begin{cases}b^2c^2+c^2a^2-a^2b^2=-abc^2\\b^2c^2+a^2b^2-a^2c^2=2ab^2c\\a^2c^2+a^2b^2-b^2c^2=2a^2bc\end{cases}}\)

\(\Rightarrow E=\frac{a^2b^2c^2}{2ab^2c}+\frac{a^2b^2c^2}{-2abc^2}+\frac{a^2b^2c^2}{2a^2bc}\)

\(\Rightarrow E=\frac{ac}{2}-\frac{ab}{2}+\frac{bc}{2}=\frac{ac-ab+bc}{2}=\frac{0}{2}=0\)

Vậy \(E=0\)

Hình như sai đề :

Ta có : \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)

\(\Leftrightarrow\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=0\)

\(\Leftrightarrow\dfrac{ab+ac+bc}{abc}=0\)

\(\Leftrightarrow ab+ac+bc=0\) ( do \(a;b;c\ne0\) ) ( 1 )

Từ ( 1 ) \(\Rightarrow ab+bc=-ac\)

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

\(\Rightarrow a^2b^2+b^2c^2+2ab^2c=a^2c^2\) ( * )

CMTT , ta được : \(\left\{{}\begin{matrix}b^2c^2+c^2a^2+2bc^2a=a^2b^2\\c^2a^2+a^2b^2+2a^2cb=b^2c^2\end{matrix}\right.\) ( *' )

Thay ( * ) và ( * ') vào E , ta được :

\(E=\dfrac{a^2b^2c^2}{a^2b^2+b^2c^2-\left(a^2b^2+b^2c^2+2b^2ac\right)}+\dfrac{a^2b^2c^2}{b^2c^2+c^2a^2-\left(b^2c^2+c^2a^2+2bc^2a\right)}\)

\(+\dfrac{a^2b^2c^2}{c^2a^2+a^2b^2-\left(c^2a^2+a^2b^2+2a^2cb\right)}\)

\(=\dfrac{a^2b^2c^2}{-2b^2ac}+\dfrac{a^2b^2c^2}{-2c^2ab}+\dfrac{a^2b^2c^2}{-2a^2cb}\)

\(=\dfrac{-ac}{2}+\dfrac{-ab}{2}+\dfrac{-bc}{2}\)

\(=\dfrac{-\left(ac+ab+bc\right)}{2}\)

\(=\dfrac{0}{2}=0\)

Vậy \(E=0\)

Áp dụng bđt Cauchy-Schwarz

\(\frac{1}{2a+b+c}=\frac{1}{\left(a+b\right)+\left(a+c\right)}\le\frac{1}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\)

\(\frac{1}{a+2b+c}=\frac{1}{\left(a+b\right)+\left(b+c\right)}\le\frac{1}{4}\left(\frac{1}{a+b}+\frac{1}{b+c}\right)\)

\(\frac{1}{a+b+2c}=\frac{1}{\left(a+c\right)+\left(b+c\right)}\le\frac{1}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\)

Cộng theo vế =>đpcm

áp dụng cái này: 
a²/x + b²/y + c²/z +d²/t ≥ (a + b +c +d)²/(x + y + z + t) (wen thuộc) 
1/a + 1/b + 1/b + 1/c ≥ 16/(a + 2b +c) 
1/a + 1/b + 1/c + 1/c ≥ 16/(a + b +2c) 
1/a + 1/a + 1/b + 1/c ≥ 16/(2a + b +c) 
Cộng 3 vế lại: 
1/a + 1/b +1/c ≥ 4[1/(a+2b+c) + 1/(b+2c+a) + 1/(c+2a+b)] 
⇔ ¼ (1/a + 1/b +1/c) ≥ 1/(a+2b+c) + 1/(b+2c+a) + 1/(c+2a+b) 
⇒ ½ (1/a + 1/b +1/c) ≥ ¼ (1/a + 1/b +1/c) ≥ 1/(a+2b+c) + 1/(b+2c+a) + 1/(c+2a+b) 
⇔ ½ (1/a + 1/b +1/c) ≥ 1/(a+2b+c) + 1/(b+2c+a) + 1/(c+2a+b) 
Dấu = xra khi a = b = c và 1/a + 1/b +1/c = 0 
⇒ dấu = không xảy ra. 
⇒ ½ (1/a + 1/b +1/c) > 1/(a+2b+c) + 1/(b+2c+a) + 1/(c+2a+b) 

Em kiểm tra lại đề ở tỉ số đầu tiên

\(\dfrac{2a+2b-2c}{c}=\dfrac{2b-2c+2a}{a}\)

Hay là: \(\dfrac{2a+2b-2c}{c}=\dfrac{2b+2c-2a}{a}\)