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a=b=c=1 suy ra Tam giác ABC là tam giác đều vì có độ dài 3 canh = nhau .
Ta có: \(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Rightarrow\frac{a}{b-c}=-\frac{b}{c-a}-\frac{c}{a-b}\)
\(\Rightarrow\frac{a}{b-c}=\frac{-b\left(a-b\right)-c\left(c-a\right)}{\left(c-a\right)\left(a-b\right)}\)
\(\Rightarrow\frac{a}{b-c}=\frac{-ab+b^2-c^2+ac}{\left(c-a\right)\left(a-b\right)}\)
\(\Rightarrow\frac{a}{\left(b-c\right)^2}=\frac{-ab+b^2-c^2+ac}{\left(c-a\right)\left(a-b\right)\left(b-c\right)}\)
Tương tự ta có: \(\frac{b}{\left(c-a\right)^2}=\frac{-bc+c^2-a^2+ab}{\left(c-a\right)\left(a-b\right)\left(b-c\right)}\)
\(\frac{c}{\left(a-b\right)^2}=\frac{-ca+a^2-b^2+bc}{\left(c-a\right)\left(a-b\right)\left(b-c\right)}\)
Cộng các đẳng thức trên ta được:
\(\frac{a}{\left(b-c\right)^2}\)\(+\frac{b}{\left(c-a\right)^2}\)\(+\frac{c}{\left(a-b\right)^2}=\)\(\frac{-ab+b^2-c^2+ac-bc+c^2-a^2+ba-ca+a^2-b^2+bc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
Vậy \(\frac{a}{\left(b-c\right)^2}\)\(+\frac{b}{\left(c-a\right)^2}\)\(+\frac{c}{\left(a-b\right)^2}=\)0 (đpcm)
Câu hỏi của Jungkookie - Toán lớp 7 - Học toán với OnlineMath
\(abc=1\Rightarrow\left(abc\right)^2=a^2b^2c^2=1\Rightarrow a^2=\frac{1}{b^2c^2}\Rightarrow\frac{1}{a^3\left(b+c\right)}=\frac{b^2c^2}{a\left(b+c\right)}=\frac{\left(bc\right)^2}{ab+ac}\)
Chứng minh tương tự ta có: \(\frac{1}{b^3\left(c+a\right)}=\frac{\left(ca\right)^2}{bc+ba};\frac{1}{c^3\left(a+b\right)}=\frac{\left(ab\right)^2}{ca+cb}\)
=> \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}=\frac{\left(ab\right)^2}{bc+ca}+\frac{\left(bc\right)^2}{ab+ca}+\frac{\left(ca\right)^2}{ab+bc}\)
Áp dụng bđt Cauchy-Schwarz dạng Engel: \(\frac{\left(ab\right)^2}{bc+ca}+\frac{\left(bc\right)^2}{ab+ca}+\frac{\left(ca\right)^2}{ab+bc}\ge\frac{\left(ab+bc+ca\right)^2}{bc+ca+ab+ca+ab+bc}=\frac{ab+bc+ca}{2}\)
Tiếp tục áp dụng bđt Cauchy với 3 số dương ta được: \(\frac{ab+bc+ca}{2}\ge\frac{3\sqrt[3]{ab.bc.ca}}{2}=\frac{3\sqrt[3]{\left(abc\right)^2}}{2}=\frac{3\sqrt[3]{1}}{2}=\frac{3}{2}\)
=> \(\frac{\left(ab\right)^2}{bc+ca}+\frac{\left(bc\right)^2}{ab+ca}+\frac{\left(ca\right)^2}{ab+bc}\ge\frac{ab+bc+ca}{2}\ge\frac{3}{2}\)
Đặt: \(A=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\)
\(\Leftrightarrow A=\frac{ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)}{abc}\)
\(\Leftrightarrow abc\cdot A=ab\left(a-b\right)+bc\left[\left(b-a\right)+\left(a-c\right)\right]+ca\left(c-a\right)\)
\(\Leftrightarrow abc\cdot A=ab\left(a-b\right)-bc\left(a-b\right)-bc\left(c-a\right)+ca\left(c-a\right)\)
\(\Leftrightarrow abc\cdot A=b\left(a-b\right)\left(a-c\right)+c\left(c-a\right)\left(a-b\right)\)
\(\Leftrightarrow abc\cdot A=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
\(\Rightarrow A=-\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{abc}\) (1)
Đặt: \(B=\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\)
\(\Leftrightarrow B=\frac{a\left(a-b\right)\left(c-a\right)+b\left(a-b\right)\left(b-c\right)+c\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(\Leftrightarrow\left(a-b\right)\left(b-c\right)\left(c-a\right)\cdot B=a\left(a-b\right)\left(c-a\right)-\left(c+a\right)\left(a-b\right)\left(b-c\right)+c\left(b-c\right)\left(c-a\right)\)
\(=a\left(a-b\right)\left(c-a\right)-c\left(a-b\right)\left(b-c\right)-a\left(a-b\right)\left(b-c\right)+c\left(b-c\right)\left(c-a\right)\)
\(=a\left(a-b\right)\left(2c-a-b\right)+c\left(b-c\right)\left(b+c-2a\right)\)
\(=3ac\left(a-b\right)-3ac\left(b-c\right)\)
\(=3ac\left(a+c-2b\right)=-9abc\)
\(\Rightarrow B=-\frac{9abc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(\Rightarrow A\cdot B=9\)