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

Ta luôn có: \(\left(\frac{a+b}{a-b}+\frac{b+c}{b-c}+\frac{c+a}{c-a}\right)^2\ge0\)\(\Leftrightarrow\left(\frac{a+b}{a-b}\right)^2+\left(\frac{b+c}{b-c}\right)^2+\left(\frac{c+a}{c-a}\right)^2+2.\)\(\left(\frac{\left(a+b\right)\left(b+c\right)}{\left(a-b\right)\left(b-c\right)}+\frac{\left(b+c\right)\left(c+a\right)}{\left(b-c\right)\left(c-a\right)}+\frac{\left(c+a\right)\left(a+b\right)}{\left(c-a\right)\left(a-b\right)}\right)\ge0\)\(\Leftrightarrow\left(\frac{a+b}{a-b}\right)^2+\left(\frac{b+c}{b-c}\right)^2+\left(\frac{c+a}{c-a}\right)^2\ge2\)(*)\(\Leftrightarrow\left(\frac{a+b}{a-b}\right)^2+1+\left(\frac{b+c}{b-c}\right)^2+1+\left(\frac{c+a}{c-a}\right)^2+1\ge5\)

\(\Leftrightarrow\frac{2\left(a^2+b^2\right)}{\left(a-b\right)^2}+\frac{2\left(b^2+c^2\right)}{\left(b-c\right)^2}+\frac{2\left(c^2+a^2\right)}{\left(c-a\right)^2}\ge5\)\(\Leftrightarrow\frac{a^2+b^2}{\left(a-b\right)^2}+\frac{b^2+c^2}{\left(b-c\right)^2}+\frac{c^2+a^2}{\left(c-a\right)^2}\ge\frac{5}{2}\)(1)

(*)\(\Leftrightarrow\left(\frac{a+b}{a-b}\right)^2-1+\left(\frac{b+c}{b-c}\right)^2-1+\left(\frac{c+a}{c-a}\right)^2-1\ge-1\)\(\Leftrightarrow\frac{4ab}{\left(a-b\right)^2}+\frac{4bc}{\left(b-c\right)^2}+\frac{4ca}{\left(c-a\right)^2}\ge-1\)\(\Leftrightarrow\frac{ab}{\left(a-b\right)^2}+\frac{bc}{\left(b-c\right)^2}+\frac{ca}{\left(c-a\right)^2}\ge-\frac{1}{4}\)(2)

Lấy (1) + (2), ta được: \(\frac{a^2+ab+b^2}{\left(a-b\right)^2}+\frac{b^2+bc+c^2}{\left(b-c\right)^2}+\frac{c^2+ca+a^2}{\left(c-a\right)^2}\ge\frac{9}{4}\)

\(\Leftrightarrow\frac{a^3-b^3}{\left(a-b\right)^3}+\frac{b^3-c^3}{\left(b-c\right)^3}+\frac{c^3-a^3}{\left(c-a\right)^3}\ge\frac{9}{4}\)(đpcm)

Chú ý: Từ đây ta có thể biến thành một BĐT khác khó hơn: \(\frac{a^3+b^3}{\left(a-b\right)^3}+\frac{b^3+c^3}{\left(b-c\right)^3}+\frac{c^3+a^3}{\left(c-a\right)^3}\ge\frac{9}{4}\)

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