Ta có M = - a + b - b - c + a + c - a

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

= 0 + 0 + 0 - a

= - a

Vì a < 0 => - a > 0

=> M > 0

a, \(a>b\) nên \(a-b>0\)

\(c>d\) nên \(c-d>0\)

Do đó : \(a-b+c-d>0\)

\(\Leftrightarrow a+c-\left(b+d\right)>0\)

\(\Leftrightarrow a+c>b+d\)

b, \(a>b>0\)nên \(\frac{a}{b}>1\)

\(c>d>0\)nên \(\frac{c}{d}>1\)

\(\Rightarrow\frac{a}{b}.\frac{c}{d}>1\)

\(\Leftrightarrow\frac{ac}{bd}>1\)

\(\Leftrightarrow ac>bd\)

ta có (a+b)(b+c)(c+a)+abc

=abc+b²c+ac²+bc²+a²b+ab²+a²c+abc+abc

=(abc+b²c+ab²)+(abc+ac²+bc²)+(abc+a²c+a²b)

=b(a+b+c)+c(a+b+c)+a(a+b+c)=0

|x|<a

nên \(x^2< a^2\)

hay -a<x<a

Có: a > b

\(\Rightarrow\) ac > bc

\(\Rightarrow\) ab + ac > ab + bc

\(\Rightarrow\) a( b + c) > b(a + c)

\(\Rightarrow\dfrac{a}{b}>\dfrac{a+c}{b+c}\)

\(a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)\cdot c+c^2\right]-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

Khi đó xảy ra 2 trường hợp:

TH1:\(a+b+c=0\Rightarrowđpcm\)

\(TH2:a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow a^2+b^2+c^2=ab+bc+ca\)

Áp dụng BĐT Bunhiacopski ta có:

\(\left(1^2+1^2+1^2\right)\left(a^2+b^2+c^2\right)\ge\left(1\cdot a+1\cdot b+1\cdot c\right)^2\)

\(\Rightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)

\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)

Dấu "=" xảy ra tại a=b=c

Vậy \(a^3+b^3+c^3=3abc\) thì \(a+b+c=0\) hoặc \(a=b=c\)