BĐT đã cho tương đương với:

\(\left(\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}\right)^2-2\left[\dfrac{ab}{\left(b-c\right)\left(c-a\right)}+\dfrac{bc}{\left(c-a\right)\left(a-b\right)}+\dfrac{ca}{\left(a-b\right)\left(b-c\right)}\right]\ge2\left(\cdot\right)\).

Mặt khác ta có: \(\dfrac{ab}{\left(b-c\right)\left(c-a\right)}+\dfrac{bc}{\left(c-a\right)\left(a-b\right)}+\dfrac{ca}{\left(a-b\right)\left(b-c\right)}=\dfrac{ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{-\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=-1\).

Do đó \(\left(\cdot\right)\Leftrightarrow\left(\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}\right)^2\ge0\) (luôn đúng).

BĐT đã cho dc c/m.

Trước hết ta có:

\(\dfrac{ab}{\left(b-c\right)\left(c-a\right)}+\dfrac{ac}{\left(b-c\right)\left(a-b\right)}+\dfrac{bc}{\left(c-a\right)\left(a-b\right)}\)

\(=\dfrac{ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(=\dfrac{ab\left(a-b\right)+b^2c-a^2c+ac^2-bc^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(=\dfrac{ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(=\dfrac{\left(a-b\right)\left(ab-ac-bc+c^2\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{\left(a-b\right)\left[a\left(b-c\right)-c\left(b-c\right)\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

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

Do đó:

\(\left(\dfrac{a}{b-c}\right)^2+\left(\dfrac{b}{c-a}\right)^2+\left(\dfrac{c}{a-b}\right)^2-2+2\)

\(=\left(\dfrac{a}{b-c}\right)^2+\left(\dfrac{b}{c-a}\right)^2+\left(\dfrac{c}{a-b}\right)^2+2\left(\dfrac{ab}{\left(b-c\right)\left(c-a\right)}+\dfrac{ac}{\left(a-b\right)\left(b-c\right)}+\dfrac{bc}{\left(c-a\right)\left(a-b\right)}\right)+2\)

\(=\left(\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}\right)^2+2\ge2\) (đpcm)

Ta có: \(a^2-b=b^2-c\Leftrightarrow a^2-b^2=b-c\)

\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=b-c\Rightarrow a+b=\frac{b-c}{a-b}\)

Tương tự CM được: \(b+c=\frac{c-a}{b-c}\) và \(c+a=\frac{a-b}{c-a}\)

Khi đó:

\(\left(a+b+1\right)\left(b+c+1\right)\left(c+a+1\right)\)

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

\(=\frac{c-b}{c-a}\cdot\frac{b-a}{b-c}\cdot\frac{a-c}{a-b}=-1\)

Vì a² - b = b² - c = c² - a

Ta có a² - b = b² - c

=> (a - b)(a + b) = b - c

=> a + b + 1 = \(\frac{a-c}{a-b}\)

Tương tự ta có : b + c + 1 = \(\frac{b-a}{b-c}\)

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

Khi đó (a + b + 1)(b + c + 1)(a + c + 1) = \(\frac{a-c}{a-b}.\frac{b-a}{b-c}.\frac{b-c}{a-c}=-1\)(đpcm) 

Ta có:\(a^2-b=b^2-c\)

\(\Leftrightarrow a^2-b^2=b-c\)

\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=b-c\)

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

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

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

Cmtt ta có:

\(\hept{\begin{cases}b^2-c=c^2-a\Leftrightarrow b+c+1=\frac{b-a}{b-c}\\c^2-a=a^2-b\Leftrightarrow c+a+1=\frac{c-b}{c-a}\end{cases}}\)

\(\Rightarrow\left(a+b+1\right)\left(b+c+1\right)\left(c+a+1\right)=\frac{a-c}{a-b}.\frac{b-c}{b-a}.\frac{c-b}{c-a}=-1\)

Cre:mạng 

 do in you free time

Đặt \(x=\frac{a}{b-c};y=\frac{b}{c-a};z=\frac{c}{a-b}\)

\(\Rightarrow xy+yz+zx=\frac{ab}{\left(b-c\right)\left(c-a\right)}+\frac{bc}{\left(c-a\right)\left(a-b\right)}+\frac{ca}{\left(a-b\right)\left(b-c\right)}=-1\) (Tự CM)  

Ta có: \(VT=x^2+y^2+z^2=\left(x+y+z\right)^2-2\left(xy+yz+zx\right)\ge2\) 

=> ĐPCM