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\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+c^3=\dfrac{3}{abc}\)
\(\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3+\dfrac{1}{c^3}-\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-\dfrac{3}{abc}=0\)
\(\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2-\dfrac{1}{c}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\dfrac{1}{c^2}\right)-\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=0\)
\(\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}-\dfrac{1}{ab}-\dfrac{1}{bc}-\dfrac{1}{ca}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b=c\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\end{matrix}\right.\)
Đề bài thiếu, cần thêm dữ liệu "a;b;c phân biệt"
Khi đó \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow ab+bc+ca=0\)
\(\Rightarrow\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)=a^2+b^2+c^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a+b+c=0\) nên trong 3 số a;b;c phải có ít nhất 1 số dương
Do vai trò của 3 biến như nhau, ko mất tính tổng quát, giả sử \(c>0\)
\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)
\(a^3+b^3+c^3=a^3+b^3+3ab\left(a+b\right)+c^3-3ab\left(a+b\right)\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)=\left(-c\right)^3+c^3-3ab\left(-c\right)=3abc=-6\)
\(\Rightarrow F=\dfrac{ab+bc+ca-\left(a^2+b^2+c^2\right)}{-6}=\dfrac{3\left(ab+bc+ca\right)}{-6}=\dfrac{ab+bc+ca}{-2}\)
\(=\dfrac{-\dfrac{2}{c}+c\left(a+b\right)}{-2}=\dfrac{-\dfrac{2}{c}+c\left(-c\right)}{-2}=\dfrac{c^2}{2}+\dfrac{1}{c}=\dfrac{c^2}{2}+\dfrac{1}{2c}+\dfrac{1}{2c}\ge3\sqrt[3]{\dfrac{c^2}{8c^2}}=\dfrac{3}{2}\)
\(F_{min}=\dfrac{3}{2}\) khi \(\left(a;b;c\right)=\left(-2;1;1\right)\) và các hoán vị
![](https://rs.olm.vn/images/avt/0.png?1311)
Có : a + b + c = 0
=> (a + b)5 = (-c)5
a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 = -c5
a5 + b5 + c5 = -5a4b - 10a3b2 - 10a2b3 - 5ab4
a5 + b5 + c5 = -5ab(a3 + 2a2b + 2ab2 + b3)
a5 + b5 + c5 = -5ab[(a3 + b3) + (2a2b + 2ab2)]
a5 + b5 + c5 = -5ab[(a + b)(a2 - ab + b2) + 2ab(a + b)]
a5 + b5 + c5 = -5ab(a + b)(a2 + b2 + ab)
a5 + b5 + c5 = 5abc(a2 + b2 + ab) (do a+b+c=0=> a+b=-c)
2(a5 + b5 + c5) = 5abc(2a2 + 2b2 + 2ab)
2(a5 + b5 + c5) = 5abc[a2 + b2 +(a2 + 2ab + b2)]
2(a5 + b5 + c5) = 5abc[a2 + b2 + (a + b)2]
2(a5 + b5 + c5) = 5abc(a2 + b2 + c2) (do a+b=-c=> (a +b )2 = c2
\(\Leftrightarrow\) \(a^5+b^5+c^5=\dfrac{5}{2}abc\left(a^2+b^2+c^2\right)\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài này đã có ở đây:
Cho abc=1CMR\(\dfrac{a+3}{\left(a+1\right)^2}+\dfrac{b+3}{\left(b+1\right)^2}+\dfrac{c+3}{\left(c+1\right)^2}\ge3\) - Hoc24