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a: Ta có: \(a+b+c=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)
Ta có: a+b+c=0
\(\Leftrightarrow\left(a+b+c\right)^3=0\)
\(\Leftrightarrow a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\)
b: Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow a+b+c=0\)
a) \(a^3+b^3+c^3=3abc\Leftrightarrow\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc=0\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)(đúng do a+b+c = 0)
a: Ta có: a+b+c=0
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)
Ta có: a+b+c=0
\(\Leftrightarrow\left(a+b+c\right)^3=0\)
\(\Leftrightarrow a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\)
b: Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Leftrightarrow a+b+c=0\)
Bài 1:
$a^3+b^3+c^3=3abc$
$\Leftrightarrow (a+b)^3-3ab(a+b)+c^3-3abc=0$
$\Leftrightarrow [(a+b)^3+c^3]-[3ab(a+b)+3abc]=0$
$\Leftrightarrow (a+b+c)[(a+b)^2-c(a+b)+c^2]-3ab(a+b+c)=0$
$\Leftrightarrow (a+b+c)[(a+b)^2-c(a+b)+c^2-3ab]=0$
$\Leftrightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0$
$\Rightarrow a+b+c=0$ hoặc $a^2+b^2+c^2-ab-bc-ac=0$
Xét TH $a^2+b^2+c^2-ab-bc-ac=0$
$\Leftrightarrow 2(a^2+b^2+c^2)-2(ab+bc+ac)=0$
$\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0$
$\Rightarrow a-b=b-c=c-a=0$
$\Leftrightarrow a=b=c$
Vậy $a^3+b^3+c^3=3abc$ khi $a+b+c=0$ hoặc $a=b=c$
Áp dụng vào bài:
Nếu $a+b+c=0$
$A=\frac{-c}{c}+\frac{-b}{b}+\frac{-a}{a}=-1+(-1)+(-1)=-3$
Nếu $a=b=c$
$P=\frac{a+a}{a}+\frac{b+b}{b}+\frac{c+c}{c}=2+2+2=6$
Ta có:
\(a^3+b^3+c^3+d^3\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+\left(c+d\right)^3-3cd\left(c+d\right)\)
\(=-\left(c+d\right)^3+3ab\left(c+d\right)+\left(c+d\right)^3-3cd\left(c+d\right)\) (vì \(a+b=-\left(c+d\right)\))
\(=3\left(c+d\right)\left(ab-cd\right)\)
Vậy đẳng thức được chứng minh.
Do \(a+b+c=1\) nên BĐT cần chứng minh tương đương:
\(2\left(a^3+b^3+c^3\right)+3abc\ge\left(ab+bc+ca\right)\left(a+b+c\right)\)
\(\Leftrightarrow2\left(a^3+b^3+c^3\right)\ge ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)\)
Thật vậy, ta có:
\(2\left(a^3+b^3+c^3\right)=\left(a^3+b^3\right)+\left(b^3+c^3\right)+\left(c^3+a^3\right)\)
\(=\left(a+b\right)\left(a^2+b^2-ab\right)+\left(b+c\right)\left(b^2+c^2-bc\right)+\left(c+a\right)\left(c^2+a^2-ca\right)\)
\(\ge\left(a+b\right)\left(2ab-ab\right)+\left(b+c\right)\left(2bc-bc\right)+\left(c+a\right)\left(2ca-ca\right)\)
\(=ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
