\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\left(a+b+c\right)\dfrac{9}{a+b+c}=9\)

Ta có :

\(M>N:\hept{\begin{cases}M=a+b-1\\N=b+c-1\end{cases}}\)

M=(a+b)-1 ; N=(b+c)-1

=> a+b > b+c

<=> b=b => a>c

=> a-c > 0

M-N= a+b-1-(b-c-1)
     = a+b-1-b-c+1
     = a+(b-b)+(-1+1)-c
    = a-c
=> M>N; M-N=a-c=> a-c>0

Ta có:

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

Thay a=1

=>\(1+\dfrac{1}{b}=b+\dfrac{1}{c}=c+1\)

*Lấy \(1+\dfrac{1}{b}=c+1\Rightarrow\dfrac{1}{b}=c\Rightarrow b=\dfrac{1}{c}\)

=>\(1+\dfrac{1}{b}=\dfrac{2}{c}=c+1\)

*Lấy \(\dfrac{2}{c}=\dfrac{c+1}{1}\)

=> 2=c(c+1)

<=> 2=c²+c

=>c=-2

*Lấy \(1+\dfrac{1}{b}=\dfrac{2}{c}\)

Thay c=-2 và quy đồng

=>\(\dfrac{b+1}{b}=-1\)

=>b+1=-b

=> b+b=-1

=>2b=-1

=> b=-1/2

Vậy b=\(-\dfrac{1}{2};c=-2\)

A=[(a+b)/a][(b+c)/b][(c+a)/c]

a+b+c=0=>a+b=-c;b+c=-a;c+a=-b

A=-(abc)/(abc)=-1

\(1=\left(a+b+c\right)^2\ge4a\left(b+c\right)\)

\(\Rightarrow b+c=\left(b+c\right).1\ge4a\left(b+c\right)\left(b+c\right)=4a\left(b+c\right)^2\ge4a.4bc=16abc\) (đpcm)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a+b+c=1\\a=b+c\\b=c\end{matrix}\right.\) \(\Rightarrow\left(a;b;c\right)=\left(\dfrac{1}{2};\dfrac{1}{4};\dfrac{1}{4}\right)\)

\(1+\dfrac{4}{b}\) hay là \(1+\dfrac{a}{b}\) vậy bạn

\(A=\left(a+\frac{1}{a}-2\right)+\left(b+\frac{1}{b}-2\right)+\left(c+\frac{1}{c}-2\right)-\left(a+b+c\right)+6\)

\(A=\frac{a^2-2a+1}{a}+\frac{b^2-2b+1}{b}+\frac{c^2-2c+1}{c}-3+6\)

\(A=\frac{\left(a-1\right)^2}{a}+\frac{\left(b-1\right)^2}{b}+\frac{\left(c-1\right)^2}{c}+3\) \(\ge3\forall a,b,c>0\)

A = 3 \(\Leftrightarrow a=b=c=1\)

Vậy min A = 3 \(\Leftrightarrow a=b=c=1\)

\(3A=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)\ge9\) (bđt AM-GM)

\(\Rightarrow3A\ge9\Leftrightarrow A\ge3\)

\("="\Leftrightarrow a=b=c=1\)