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\(a^3+b^3+c^3=3abc\\ \Rightarrow a^3+b^3+c^3-3abc=0\\ \Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\\ \Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{matrix}\right.\)
\(\Rightarrow a^2+b^2+c^2=ab+bc+ac\left(a+b+c\ne0\right)\\ \Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\\ \Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\\ \Rightarrow a=b=c\\ \Rightarrow B=\dfrac{2}{a}.\dfrac{2}{b}.\dfrac{2}{c}=\dfrac{8}{abc}\)
\(\text{Ta có: }\)\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=a^2+b^2+c^2\)
\(\Rightarrow ab+bc+ca=0\Rightarrow-ab=bc+ca\)
\(VT=\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{b^3c^3+a^3b^3+a^3c^3}{\left(abc\right)^3}\)
\(=\dfrac{\left(bc+ca\right)^3-3abc^2\left(bc+ca\right)+\left(ab\right)^3}{\left(abc\right)^3}\)
\(=\dfrac{\left(-ab\right)^3+3\left(abc\right)^2+\left(ab\right)^3}{\left(abc\right)^3}=\dfrac{\left[-\left(ab\right)^3+\left(ab\right)^3+3\left(abc\right)^2\right]}{\left(abc\right)^3}\)
\(=\dfrac{3\left(abc\right)^2}{\left(abc\right)^3}=\dfrac{3}{abc}=VP\)
Bạn tham khảo tại đây:
Câu hỏi của Hoàng Tuấn - Toán lớp 8 | Học trực tuyến
Ta có:
\(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(c+a\right)}+\dfrac{1}{c^3\left(a+b\right)}\ge\dfrac{3}{2}\)
\(\dfrac{1^2}{a^3\left(b+c\right)}+\dfrac{1^2}{b^3\left(c+a\right)}+\dfrac{1^2}{c^3\left(a+b\right)}\ge\dfrac{3}{2}\)
\(\dfrac{a^2b^2c^2}{a^3\left(b+c\right)}+\dfrac{a^2b^2c^2}{b^3\left(c+a\right)}+\dfrac{a^2b^2c^2}{c^3\left(a+b\right)}\ge\dfrac{3}{2}\)
\(\dfrac{b^2c^2}{a\left(c+b\right)}+\dfrac{a^2c^2}{b\left(c+a\right)}+\dfrac{a^2b^2}{c\left(a+b\right)}\ge\dfrac{3}{2}\)
Áp dụng BĐT Svacxo ta có:
\(\dfrac{b^2c^2}{a\left(b+c\right)}+\dfrac{a^2c^2}{b\left(c+a\right)}+\dfrac{a^2b^2}{c\left(a+b\right)}\ge\dfrac{\left(ab+bc+ca\right)^2}{a\left(b+c\right)+b\left(a+c\right)+c\left(a+b\right)}\) \(\dfrac{b^2c^2}{a\left(b+c\right)}+\dfrac{a^2c^2}{b\left(c+a\right)}+\dfrac{a^2b^2}{c\left(a+b\right)}\ge\dfrac{\left(ab+bc+ca\right)}{2}\) (1)
Chứng minh: \(\dfrac{ab+bc+ca}{2}\ge\dfrac{3}{2}\Leftrightarrow ab+bc+ca\ge3\)
Áp dụng BĐT Cosi ta có:
\(ab+bc+ca\ge3\sqrt[3]{ab.bc.ca}\)
\(ab+bc+ca\ge3\) (2)
Từ (1) và (2)
=> ĐPCM
Lời giải:
Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow \frac{ab+bc+ac}{abc}=0\Leftrightarrow ab+bc+ac=0\)
\(\Leftrightarrow 2(ab+bc+ac)=0\)
Cộng cả hai vế với \(a^2+b^2+c^2\) thì:
\(a^2+b^2+c^2+2(ab+bc+ac)=a^2+b^2+c^2\)
\(\Leftrightarrow (a+b+c)^2=a^2+b^2+c^2\)
Do đó ta có đpcm.
3) Biến đổi tương đương:
\(8\left(a^3+b^3+c^3\right)\ge\left(a+b\right)^3+\left(b+c\right)^3+\left(a+c\right)^3\) (1)
\(\Leftrightarrow\left(a^3+b^3\right)+\left(b^3+c^3\right)+\left(a^3+c^3\right)+6\left(a^3+c^3+b^3\right)\)
\(\ge\left(a^3+b^3\right)+\left(b^3+c^3\right)+\left(a^3+c^3\right)+3ab\left(a+b\right)+3bc\left(b+c\right)+3ac\left(a+c\right)\)
\(\Leftrightarrow2\left(a^3+b^3+c^3\right)\ge ab\left(a+b\right)+bc\left(b+c\right)+ac\left(a+c\right)\)
\(\Leftrightarrow\left[a^3+b^3-ab\left(a+b\right)\right]+\left[a^3+c^3-ac\left(a+c\right)\right]+\left[b^3+c^3-bc\left(b+c\right)\right]\ge0\)
\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2+\left(a+c\right)\left(a-c\right)^2+\left(b+c\right)\left(b-c\right)^2\ge0\) luôn đúng do a, b, c > 0
=> (1) đúng
Dấu "=" xảy ra khi a = b = c
4) Ta có: a+b>c ; b+c>a; a+c>b
Xét \(\dfrac{1}{a+c}+\dfrac{1}{b+c}>\dfrac{1}{a+b+c}+\dfrac{1}{b+c+a}=\dfrac{2}{a+b+c}>\dfrac{2}{a+b+a+b}=\dfrac{1}{a+b}\)
Tương tự: \(\dfrac{1}{a+b}+\dfrac{1}{a+c}>\dfrac{1}{b+c}\)
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}>\dfrac{1}{a+c}\)
Vậy suy ra được điều phải chứng minh
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{abc}\Rightarrow ab+bc+ac=1\)
Ta có \(1+a^2=a^2+ab+bc+ac=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)
Tương tự ta được \(1+b^2=\left(a+b\right)\left(b+c\right)\); \(1+c^2=\left(a+c\right)\left(b+c\right)\)
\(\Rightarrow A=\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)\)
\(\Rightarrow A=\left[\left(a+b\right)\left(a+c\right)\left(b+c\right)\right]^2\) \(\Rightarrow A\) là số chính phương
\(\left(a+b+c\right)^2=a^2+b^2+c^2\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)
\(\Leftrightarrow ab+bc+ac=0\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{a}=x\\\dfrac{1}{b}=y\\\dfrac{1}{c}=z\end{matrix}\right.\) \(\Rightarrow x+y+z=0\) \(\Rightarrow z=-\left(x+y\right)\)
Đẳng thức cần chứng minh: \(x^3+y^3+z^3=3xyz\) với \(x+y+z=0\)
Ta có:
\(x^3+y^3+z^3=x^3+y^3-\left(x+y\right)^3=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)^3\)
\(=\left(x+y\right)\left(x^2-xy+y^2-\left(x+y\right)^2\right)=\left(x+y\right)\left(-3xy\right)\)
\(=-\left(x+y\right).3xy=z.3xy=3xyz\)
Vậy \(x^3+y^3+z^3=3xyz\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
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