từ giả thiết suy ra : 

a²b  - a³bc - b²c + ab²c² = ab² - ab³c - a²c + a²bc²

\(\Rightarrow\)ab ( a - b ) + c ( a² - b² ) = abc² ( a - b ) + abc ( a² - b² )

\(\Rightarrow\)( a - b ) ( ab + ac + bc ) = abc ( a - b ) ( c + a + b )

chia 2 vế cho abc ( a - b ) \(\ne\)0 

Ta có :\(\frac{a}{b}=\frac{b}{c}\)

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

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

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

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

=> \(\frac{a}{c}=\frac{a^2+b^2}{b^2+c^2}\left(\text{đpcm}\right)\)

Cho xem đáp án nhé

Có \(a^2+ab+\frac{b^2}{3}=c^2+\frac{b^2}{3}+a^2+ac+c^2\left(=25\right)\)

\(\Rightarrow a^2+ab+\frac{b^2}{3}=2c^2+\frac{b^2}{3}+a^2+ac\\ \Rightarrow ab=2c^2+ac\\ \Rightarrow ab+ac=2c^2+2ac\\ \Rightarrow a\left(b+c\right)=2c\left(a+c\right)\\ \Rightarrow\frac{2c}{a}=\frac{b+c}{a+c}\)

a/

\(=\frac{a+b}{b^2}.\frac{\left|a\right|.b^2}{\left|a+b\right|}=\frac{\left(a+b\right).b^2.\left|a\right|}{b^2\left(a+b\right)}=\left|a\right|\)

b/

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}-\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\frac{2\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)

\(\frac{a^2}{b^2}+\frac{b^2}{c^2}\ge2\sqrt{\frac{a^2b^2}{b^2c^2}}=2\left|\frac{a}{c}\right|\ge\frac{2a}{c}\)

Tương tự: \(\frac{a^2}{b^2}+\frac{c^2}{a^2}\ge\frac{2c}{b}\) ; \(\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{2b}{a}\)

Cộng vế với vế:

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

Dấu "=" xảy ra khi \(a=b=c\)

Cảm ơn bạn.

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

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

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

Ta có:

\(\frac{a^2}{b^2}+1\ge2.\frac{a}{b}\)

\(\frac{b^2}{c^2}+1\ge2.\frac{b}{c}\)

\(\frac{c^2}{a^2}+1\ge2.\frac{c}{a}\)

Cộng vế theo vế ta được

\(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+3\ge2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)

\(\Leftrightarrow\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-3\)

\(\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3\sqrt{\frac{a}{b}.\frac{b}{c}.\frac{c}{a}}-3=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\)
Dấu =  xảy ra khi a = b = c

Ta co: \(\frac{a^2}{b^2}\ge\frac{a}{b}\); \(\frac{b^2}{c^2}\ge\frac{b}{c}\);\(\frac{c^2}{a^2}\ge\frac{c}{a}\)\(\Rightarrow dpcm\)

\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2\left(a+b+c\right)}{abc}}\) ( do \(a+b+c=0\) )

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

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

\(=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\) ( đpcm )

Từ a/b = b/c 

Suy ra : bb = ac 

b² = ac 

vậy : a² + b² / b²+ c² = a² + ac / ac + c² = a(a+c) / c(a+c) = a/c 

Vậy : Ta có được cái cần chứng minh :)) 
Lớp mình vừa kiểm tra 15' bài này xong . 

cảm ơn bạn