Với \(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}b+c=-a\\c+a=-b\\a+b=-c\end{matrix}\right.\)

\(B=\dfrac{a+b}{a}\cdot\dfrac{a+c}{c}\cdot\dfrac{b+c}{b}=\dfrac{-abc}{abc}=-1\)

Với \(a+b+c\ne0\)

\(\dfrac{a+b-2021c}{c}=\dfrac{b+c-2021a}{a}=\dfrac{c+a-2021b}{b}=\dfrac{-2019\left(a+b+c\right)}{a+b+c}=-2019\\ \Leftrightarrow\left\{{}\begin{matrix}a+b-2021c=-2019c\\b+c-2021a=-2019a\\c+a-2021b=-2019b\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)

\(B=\dfrac{a+b}{a}\cdot\dfrac{a+c}{c}\cdot\dfrac{b+c}{b}=\dfrac{2a\cdot2b\cdot2c}{abc}=8\)

Với a+b+c=0⇔⎧⎪⎨⎪⎩b+c=−ac+a=−ba+b=−ca+b+c=0⇔{b+c=−ac+a=−ba+b=−c

Ta có: \(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}\)\(=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

=> a+b=2c; b+c=2a; c+a=2b

Thay vào A ta được: A=((a+b)/b)((c+b)/c)((a+c)/a)

=2c/b.2a/c.2b/a=2.2.2=8

\(a+b+c=1=>\left\{{}\begin{matrix}1-a=b+c\\1-b=a+c\\1-c=a+b\\\end{matrix}\right.\)

\(=>A=\left(\dfrac{1}{a}-1\right)\left(\dfrac{1}{b}-1\right)\left(\dfrac{1}{c}-1\right)=\left(\dfrac{1-a}{a}\right)\left(\dfrac{1-b}{b}\right)\left(\dfrac{1-c}{c}\right)\)

\(=\left(\dfrac{b+c}{a}\right)\left(\dfrac{a+c}{b}\right)\left(\dfrac{a+b}{c}\right)\)

bbđt AM-GM

\(=>A\ge\dfrac{2\sqrt{bc}.2\sqrt{ac}.2\sqrt{ab}}{abc}=\dfrac{8abc}{abc}=8\left(đpcm\right)\)

dấu"=" xảy ra<=>\(a=b=c=\dfrac{1}{3}\)

Đặt vế trái BĐT cần chứng minh là P

Ta có:

\(P=\left(\dfrac{a+b+c}{a}-1\right)\left(\dfrac{a+b+c}{b}-1\right)\left(\dfrac{a+b+c}{c}-1\right)\)

\(P=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\ge\dfrac{2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}}{abc}=8\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)

TH1: \(a+b+c=0\Rightarrow P=\dfrac{\left(-c\right).\left(-a\right).\left(-b\right)}{abc}=-1\)

TH2: \(a+b+c\ne0\)

\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{a+c-b}{b}=\dfrac{a+b+c}{a+b+c}=1\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\) 

\(\Rightarrow P=\dfrac{2a.2b.2c}{abc}=8\)

1.

Đặt \(\left(x;y;z\right)=\left(\dfrac{a}{a+b};\dfrac{b}{b+c};\dfrac{c}{c+a}\right)\Rightarrow\left\{{}\begin{matrix}1-x=\dfrac{b}{b+a}\\1-y=\dfrac{c}{b+c}\\1-z=\dfrac{a}{a+c}\end{matrix}\right.\)

\(\Rightarrow xyz=\dfrac{1}{8}\\ xyz=\left(1-x\right)\left(1-y\right)\left(1-z\right)\\ \Rightarrow xyz=1-\left(x+y+z\right)+\left(xy+yz+zx\right)-xyz\\ \Rightarrow2xyz=1-\left(x+y+z\right)+\left(xy+yz+zx\right)=\dfrac{1}{4}\\ \Rightarrow x+y+z=\dfrac{3}{4}+xy+yz+zx\)

\(\RightarrowĐpcm\)

2.

Áp dụng t/c dtsbn:

\(\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}=\dfrac{a+b+c}{a+b+c}=1\\ \Rightarrow\left\{{}\begin{matrix}a+b-c=c\\a+c-b=b\\b+c-a=a\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\Rightarrow a=b=c\)

\(\Rightarrow P=\dfrac{\left(a+a\right)\left(a+a\right)\left(a+a\right)}{a\cdot a\cdot a}=\dfrac{8a^3}{a^3}=8\)

\(\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}=\dfrac{a+b-c+a+c-b+b+c-a}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)

\(\Rightarrow\left\{{}\begin{matrix}a+b-c=c\\a+c-b=b\\b+c-a=a\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\)

\(P=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{2a.2b.2c}{abc}=8\)

Áp dụng tính chất dãy tỉ số bằng nhau:

(a + b - c)/c = (a + c - b)/b = (b + c - a)/a = (a + b - c + a + c - b + b + c - a)/(a + b + c) = 1

--> a + b - c = c

a + c - b = b

b + c - a = a

--> a + b = 2c

a + c = 2b

b + c = 2a

Ta có: P = (a + b)(b + c)(a + c)/(abc) = 2c.2a.2b/(abc) = 8

Áp dụng t/c dtsbn ta có:

\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}=\dfrac{a+b+c}{a+b+c}=1\)

\(\dfrac{a+b-c}{c}=1\Rightarrow a+b-c=c\Rightarrow a+b=2c\\ \dfrac{b+c-a}{a}=1\Rightarrow b+c-a=a\Rightarrow b+c=2a\\ \dfrac{c+a-b}{b}=1\Rightarrow c+a-b=b\Rightarrow c+a=2b\)

\(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)\\ =\dfrac{\left(a+b\right)\left(a+c\right)\left(b+c\right)}{abc}\\ =\dfrac{2c.2b.2a}{abc}\\ =\dfrac{8abc}{abc}\\ =8\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được: 

\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}=\dfrac{a+b+c}{a+b+c}=1\)

Do đó: 

\(\left\{{}\begin{matrix}a+b-c=c\\b+c-a=a\\c+a-b=b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)

Thay a+b=2c;b+c=2a và c+a=2b vào biểu thức \(P=\dfrac{\left(a+b\right)\left(b+c\right)\left(a+b\right)}{abc}\), ta được: 

\(P=\dfrac{2a\cdot2b\cdot2c}{abc}=\dfrac{8abc}{abc}=8\)

Vậy: P=8

Ta có: \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\) = \(\dfrac{a+b-c+b+c-a+c+a-b}{a+b+c}\) (t/c dãy tỉ số bằng nhau)

hay \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=1\) (1)

Ta cũng có: \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{a+b-c+b+c-a}{a+c}\) (t/c dãy tỉ số bằng nhau)

hay \(\dfrac{a+b-c}{c}=\dfrac{2b}{a+c}\) (2)

Từ (1) và (2) \(\Rightarrow\) \(\dfrac{2b}{a+c}=1\) \(\Leftrightarrow\) a + c = 2b (*)

Tương tự ta cũng có: a + b = 2c (**); b + c = 2a (***)

Thay (*); (**); (***) vào P ta được:

P = \(\dfrac{2a.2b.2c}{abc}\) = 2.2.2 = 8

Vậy P = 8

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