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\(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\c+a=-b\\b+c=-a\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}A=a.\left(-c\right).\left(-b\right)=abc\\B=b.\left(-a\right).\left(-c\right)=abc\\C=c.\left(-b\right).\left(-a\right)=abc\end{matrix}\right.\)
\(\Rightarrow A=B=C\)
\(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge a+b+c\)
\(\Leftrightarrow\dfrac{a^2c+b^2a+c^2b}{abc}\ge a+b+c\)
\(\Leftrightarrow a^2c+b^2c+c^2b\ge abc\left(a+b+c\right)\)
\(\Leftrightarrow a^2c+b^2c+c^2b\ge a^2bc+ab^2c+abc^2\)
\(\Leftrightarrow a^2c+b^2c+c^2b-a^2bc-ab^2c-abc^2\ge0\)
\(\Leftrightarrow a^2c\left(1-b\right)+b^2c\left(1-a\right)+c^2b\left(1-c\right)\ge0\)
-Sửa đề: \(0< a,b,c\le1\) thì BĐT mới đúng.
a/
\(a^2+b^2+c^2+29ab+bc+ca=3\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\) \(\Rightarrow a=b=c\)
b/ \(a^3+b^3+c^3=\left(a+b\right)^3+c^3-3ab\left(a+b\right)\)
\(=\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2\right)-3ab\left(a+b\right)\)
\(=-3ab\left(a+b\right)=-3ab\left(-c\right)=3abc\)
c/ Không, vì \(a=b=c\ne\) thì \(a^3+b^3+c^3=3a^3=3abc\) vẫn đúng
Đặt x=b-c; y=c-a; z=a-b
=>x+y+z=b-c+c-a+a-b=0
Ta có: \(\frac{a}{b-c} + \frac{b}{c-a} + \frac{c}{a-b} = 0\)
=>\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)
=>ayz+bxz+cxy=0
=>\(a=\frac{(y-z)x^2}{y^2 + yz + z^2};\quad b=\frac{(z-x)y^2}{z^2 + zx + x^2};\quad c=\frac{(x-y)z^2}{x^2 + xy + y^2}\)
A=\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=\frac{a}{x^2}+\frac{b}{y^2}+\frac{c}{z^2}\)
\(=\frac{1}{x^2}\cdot\frac{(y-z)x^2}{y^2 + yz + z^2}+\frac{1}{y^2}\cdot\frac{(z-x)y^2}{z^2 + zx + x^2}+\frac{1}{z^2}\cdot\frac{(x-y)z^2}{x^2 + xy + y^2}\)
\(=\frac{y-z}{y^2 + yz + z^2}+\frac{z-x}{z^2 + zx + x^2}+\frac{x-y}{x^2 + xy + y^2}\)
x+y+z=0
=>x+y=-z; x+z=-y; y+z=-x
\(y^2 + yz + z^2 = (y + z)^2 - yz = (-x)^2 - yz = x^2 - yz\)
\(z^2 + zx + x^2 = y^2 - zx\)
\(x^2 + xy + y^2 = z^2 - xy\)
\(A=\frac{y-z}{x^2 - yz}+\frac{z-x}{y^2 - zx}+\frac{x-y}{z^2 - xy}\)
\(=\frac{(y-z)(y^2 - zx)(z^2 - xy) + (z-x)(z^2 - xy)(x^2 - yz) + (x-y)(x^2 - yz)(y^2 - zx)}{(x^2 - yz)(y^2 - zx)(z^2 - xy)}\)
=0
=>ĐPCM
Đề bài: Cho 3 số \(a+b+c=0\)..........
Vì \(a+b+c=0\)\(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)
\(\Rightarrow A=a\left(a+b\right)\left(c+a\right)=a.\left(-c\right).\left(-b\right)=abc\)(1)
\(B=b\left(b+c\right)\left(a+b\right)=b.\left(-a\right).\left(-c\right)=abc\)(2)
\(C=c\left(c+a\right)\left(b+c\right)=c.\left(-b\right).\left(-a\right)=abc\)(3)
Từ (1) , (2) và (3) \(\Rightarrow A=B=C\)
3 số mà thêm d vô mần chi rứa:v
Ta có : \(a+b+c=0< =>\hept{\begin{cases}a+b=-c\\a+c=-b\\b+c=-a\end{cases}}\)
Thay vào các biểu thức A,B,C ta có :
\(\hept{\begin{cases}A=a.\left(-c\right).\left(-b\right)=abc\\B=b.\left(-a\right).\left(-c\right)=abc\\C=c.\left(-b\right).\left(-a\right)=abc\end{cases}}\)
Suy ra \(A=B=C\)