Có \(\left(x-y\right)^2\ge0\Leftrightarrow x^2+y^2\ge2xy\)
Thay x,y lần lượt là các cặp \(\left(\frac{a}{b};\frac{b}{c}\right);\left(\frac{b}{c};\frac{c}{a}\right);\left(\frac{c}{a};\frac{a}{b}\right)\) ta được \(\frac{a^2}{b^2}+\frac{b^2}{c^2}\ge2\frac{a}{c}\)       \(\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge2\frac{b}{a}\)       \(\frac{c^2}{a^2}+\frac{a^2}{b^2}\ge2\frac{c}{b}\)
Cộng lại ta có \(2\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\ge2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\Leftrightarrow\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\)
Dấu = xảy ra khi a=b=c 

"Chấm" nhẹ hóng cao nhân ạ :)

P/s: mong các bác giải theo cách lớp 8 ạ :) Tặng 5SP / 1 câu nhé ;)

Câu 3: Tham khảo đây nhá: Câu hỏi của Trương Thanh Nhân, t làm r,giờ lười đánh lại.

tth, bài 3 làm thế chắc chết  cauchy là ra thôi 

Ta có : \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)

\(\Leftrightarrow\frac{1}{c}=\frac{a+b}{2ab}\)

\(\Leftrightarrow ca+cb=2ab\)

\(\Leftrightarrow ac-ab=ab-bc\)

\(\Leftrightarrow a\left(c-b\right)=b\left(a-c\right)\)

\(\Leftrightarrow\frac{a}{b}=\frac{a-c}{c-b}\left(đpcm\right)\)

Ta có: \(\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)\left(a+b+c\right)=1.\left(a+b+c\right)\)

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

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

=> \(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}=0\)

Ta có : \(\frac{a}{b}=\frac{b}{c}\)
\(\Rightarrow\left(\frac{a}{b}\right)^2=\left(\frac{b}{c}\right)^2=\frac{ab}{bc}\)(Áp dụng tính chất a = b => a² = b² = ab)
\(\Rightarrow\frac{a^2}{b^2}=\frac{b^2}{c^2}=\frac{ab}{bc}\)
\(\Rightarrow\frac{a^2+b^2}{b^2+c^2}=\frac{a}{c}\)(Trừ khử b trên tử và dưới mẫu còn a/c)

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

mà \(\left(\frac{a}{b}\right)^2=\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}=\frac{a}{c}\)

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

Ta có :

 \(\left(a+b+c\right)^2=a^2+b^2+c^2\)

\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=a^2+b^2+c^2\)

\(\Rightarrow2\left(ab+bc+ca\right)=0\)

\(\Rightarrow ab+bc+ca=0\)

\(\Rightarrow\frac{ab+bc+ca}{abc}=0\)

\(\Rightarrow\frac{ab}{abc}+\frac{bc}{abc}+\frac{ca}{abc}=0\)

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

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

\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)

\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab\left(\frac{1}{a}+\frac{1}{b}\right)}=-\frac{1}{c^3}\)

\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{ab\left(-\frac{1}{c}\right)}=0\)

\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}-\frac{3}{abc}=0\)

\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\) (ĐPCM)

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

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

\(\frac{1}{c}=\frac{b+a}{2ab}\)

suy ra \(2ab=c\left(b+a\right)\)

\(2ab=cb+ca\)

suy ra \(ab+ab=cb+ca\)

suy a \(ab-cb=ca-ab\)

suy ra \(b\left(a-c\right)=a\left(c-b\right)\)

suy ra \(\frac{a}{b}=\frac{a-c}{c-b}\left(Đpcm\right)\)

chua hieu may

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

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

\(\frac{1}{c}=\frac{b+a}{2ab}\)

\(\Rightarrow2ab=c\left(b+a\right)\)

\(2ab=cb+ca\)

\(\Rightarrow ab+ab=cb+ca\)

\(\Rightarrow ab-cd=ca-ab\)

\(\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)

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