Dat \(\hept{\begin{cases}A=\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\\B=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\end{cases}}\)

Ta co:\(A=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge2+2+2=6\left(1\right)\)

\(B=\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}-3\)

\(=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\ge\left(a+b+c\right).\frac{9}{2\left(a+b+c\right)}-3=\frac{3}{2}\left(2\right)\)

Cong ve voi ve cua (1) va (2) ta duoc:

\(P=A+B\ge6+\frac{3}{2}=\frac{15}{2}\)

Dau '=' xay ra khi \(a=b=c\)

Chứng minh ĐBT:\(\frac{b}{a}+\frac{a}{b}\ge2\left(a,b\ne0\right)\)(Dấu "="\(\Leftrightarrow a=b=1\))

Ta có: \(\left(a-b\right)^2\ge0\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow a^2+b^2\ge2ab\)

\(\Leftrightarrow\frac{a^2+b^2}{ab}\ge2\)

\(\Leftrightarrow\frac{a}{b}+\frac{b}{a}\ge2\left(đpcm\right)\)

Vậy \(\frac{b+c}{a}+\frac{a}{b+c}\ge2\)

\(\frac{a+c}{b}+\frac{b}{c+a}\ge2\)

\(\frac{a+b}{c}+\frac{c}{b+a}\ge2\)

\(\Rightarrow P\ge6\)

Vậy \(P_{min}=6\Leftrightarrow\hept{\begin{cases}a=b+c\\b=a+c\\c=a+b\end{cases}}\)

\(B=\frac{2001}{2}\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right].\left[\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right]\)

\(B=\frac{2001}{2}\left[\frac{a+b}{a+b}+\frac{a+b}{b+c}+\frac{a+b}{c+a}+\frac{b+c}{a+b}+\frac{b+c}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}+\frac{c+a}{b+c}+\frac{c+a}{c+a}\right]\)

\(B=\frac{2001}{2}\left[1+\left(\frac{a+b}{b+c}+\frac{b+c}{a+b}\right)+\left(\frac{a+b}{c+a}+\frac{c+a}{a+b}\right)+1+\left(\frac{b+c}{c+a}+\frac{c+a}{b+c}\right)+1\right]\)

Dễ dàng chứng minh được \(\frac{x}{y}+\frac{y}{x}\ge2\). Suy ra:

\(\left(\frac{a+b}{b+c}+\frac{b+c}{a+b}\right)\ge2\); \(\left(\frac{a+b}{c+a}+\frac{c+a}{a+b}\right)\ge2\); \(\left(\frac{b+c}{c+a}+\frac{c+a}{b+c}\right)\ge2\)

=> \(B\ge\frac{2001}{2}.\left(3+2+2+2\right)=\frac{18009}{2}\)

Dấu = khi và chỉ khi \(\frac{a+b}{b+c}=\frac{b+c}{a+b}=\frac{c+a}{b+c}\Rightarrow a=b=c\)

vậy Min B = 18009/2

Ta có : \(P=a+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\)

\(=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\)

Mặt khác \(\frac{x}{y}+\frac{y}{x}\ge2\)Với mọi \(x,y\)dương \(\Rightarrow P=3+2+2+2=9\)

Vậy \(Pmir=9\)khi \(a=b=c\)

2) Ta có :  \(\left|x-1\right|+\left|1-x\right|=2\) (1)

Xét 3 trường hợp : 

1. Với \(x>1\) , phương trình (1) trở thành : \(x-1+x-1=2\Leftrightarrow2x=4\Leftrightarrow x=2\) (thoả mãn)

2. Với \(x< 1\), phương trình (1) trở thành : \(1-x+1-x=2\Leftrightarrow2x=0\Leftrightarrow x=0\)(thoả mãn)

3. Với x = 1 , phương trình vô nghiệm.

Vậy tập nghiệm của phương trình : \(S=\left\{0;2\right\}\)

1) Cách 1:

Ta có ; \(A=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\)

\(=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)\)

Mặt khác theo bất đẳng thức Cauchy :\(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{a}{b}.\frac{b}{a}}=2\) ;\(\frac{b}{c}+\frac{c}{b}\ge2\) ; \(\frac{c}{a}+\frac{a}{c}\ge2\)

\(\Rightarrow A\ge1+2+2+2=9\). Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{a}{b}=\frac{b}{a}\\\frac{b}{c}=\frac{c}{b}\\\frac{a}{c}=\frac{c}{a}\end{cases}}\)\(\Leftrightarrow a=b=c\)

Vậy Min A = 9 <=> a = b = c

Cách 2 : Sử dụng bđt Bunhiacopxki : \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\left(1+1+1\right)^2=9\)

Khai triển của biểu thức trên là:

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

  =\(1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\)

  =\(3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\)

Mặt khác: \(\frac{x}{y}+\frac{y}{x}\ge2\)với mọi x,y dương \(\Rightarrow\frac{P}{3+2+2+2}=9\)

Vậy \(P_{min}=9\Leftrightarrow a=b=c\)

GTNN là 9 (BĐT AM-GM)