Đặt x=a + b - 2c
       y=b+c-2a
       z=c+a-2b
=>x+y+z=(a + b - 2c)+(b+c-2a)+(c+a-2b)
=>x+y+z=0
=>x+y= - z                         (1)
=>(x+y)^3=(-z)^3
=>x^3+y^3+3xy(x+y)=(-z)^3
=>x^3+y^3+z^3 +3xy(-z)=0        {vì x+y=-z [theo (1)]}
=>x^3+y^3+z^3 -3xyz=0
=>x^3+y^3+z^3 =3xyz
Vậy (a + b - 2c)^3 + (b + c - 2a)^3 + (c + a - 2b)^3=3(a + b - 2c) (b + c - 2a)(c + a - 2b)

a>b =>2a>2b =>-2a<-2b =>3-2a<3-2b. Mà 3-2b<4-2b. Vậy 3-2a<4-2b (tính chất bắc cầu).

Ta có \(a>b\Rightarrow2a>2b\Rightarrow-2a<-2b\)

Mà \(3<4\)

Do đó \(3-2a<4-2b\)

Nhận xét: \(b^3c-cb^3=0;b^2c-cb^2=0.\).Nên phân thức trở thành:

\(\frac{a^3b-ab^3+c^3a-ca^3}{a^2b-ab^2+c^2a-ca^2}=\frac{a^3\left(b-c\right)-a\left(b^3-c^3\right)}{a^2\left(b-c\right)-a\left(b^2-c^2\right)}\)
\(=\frac{a\left(b-c\right)\left\{a^2-\left(b^2-bc+c^2\right)\right\}}{a\left(b-c\right)\left\{a-\left(b+c\right)\right\}}\)
\(=\frac{a^2-\left(b^2-bc+c^2\right)}{a-\left(b+c\right)}=\frac{a^2-\left(b+c\right)^2+3bc}{a-\left(b+c\right)}\)
\(=a+b+c+\frac{3bc}{a-b-c}\).

a: \(A=\left(5xy-2xy+4xy\right)+3x-2y-y^2\)

\(=7xy+3x-2y-y^2\)

b: \(B=\left(\dfrac{1}{2}ab^2-\dfrac{7}{8}ab^2-\dfrac{1}{2}ab^2\right)+\left(\dfrac{3}{4}a^2b-\dfrac{3}{8}a^2b\right)\)

\(=\dfrac{-7}{8}ab^2+\dfrac{3}{8}a^2b\)

c: \(C=\left(2a^2b+5a^2b\right)+\left(-8b^2-3b^2\right)+\left(5c^2+4c^2\right)\)

\(=7a^2b-11b^2+9c^2\)

\(A=5xy-y^2-2xy+4xy+3x-2y\)

\(A=-y^2+7xy+3x-2y\)

\(B=\dfrac{1}{2}ab^2-\dfrac{7}{8}ab^2+\dfrac{3}{4}a^2b-\dfrac{3}{8}a^2b-\dfrac{1}{2}ab^2\)

\(B=\dfrac{3}{8}a^2b-\dfrac{7}{8}ab^2\)

\(C=2a^2b-8b^2+5a^2b+5c^2-3b^2+4c^2\)

\(C=7a^2b-11b^2+9c^2\)

\(A=7xy-y^2+3x-2y\)

\(B=\dfrac{3}{8}a^2b-\dfrac{7}{8}ab^2\)
\(C=7a^2b-11b^2+9c^2\)

A = 2a²b² + 2b²c² + 2a²c² − a⁴ − b⁴ − c⁴

<=> A = 4a²c² − ( a⁴+b⁴ + c⁴ − 2a²b² + 2a²c² − 2b²c² )

<=> A = 4a²c² − ( a² − b² + c²)²

<=> A = ( 2ac + a² − b² + c² ) ( 2ac − a² + b² − c² )

<=> A = [ (a+c)² − b² ] ( b² − (a−c)²)

<=> A = ( a+b+c) (a+c−b) (b+a−c) (b−a+c)
Mà a, b, c là 3 cạnh của tam giác nên: Mà a, b, ca, b, c là 33 cạnh của tam giác nên:\

a+b+c>0

a+c−b>0

b+a−c>0

b−a+c>

=> (a+b+c)(a+c−b)(b+a−c)(b−a+c)>0

A>0 (Dpcm)

Đặt \(\hept{\begin{cases}-a+2b+2c=x\\2a-b+2c=y\\2a+2b-c=z\end{cases}}\)\(\Rightarrow\hept{\begin{cases}a=\frac{2y+2z-x}{9}\\b=\frac{2z+2x-y}{9}\\c=\frac{2x+2y-z}{9}\end{cases}}\)

Vì a,b,c là độ dài 3 cạnh của 1 tam giác nên x,y,z>0

Khi đó : \(VT=\frac{2y+2z-x}{9x}+\frac{2z+2x-y}{9y}+\frac{2x+2y-z}{9z}\)

\(=\frac{2}{9}\left(\frac{x}{y}+\frac{y}{x}\right)+\frac{2}{9}\left(\frac{y}{z}+\frac{z}{y}\right)+\frac{2}{9}\left(\frac{z}{x}+\frac{x}{z}\right)-\frac{1}{3}\)

\(\ge\frac{2}{9}.2+\frac{2}{9}.2+\frac{2}{9}.2-\frac{1}{3}\)(BĐT Cauchy cho 2 số không âm)

\(=\frac{4}{9}.3-\frac{1}{3}=\frac{4}{3}-\frac{1}{3}=1\)

\(\frac{a}{2b+2c-a}+\frac{b}{2a+2c-b}+\frac{c}{2a+2b-c}\)

\(\frac{a^2}{2ab+2ac-a^2}+\frac{b^2}{2ab+2bc-b^2}+\frac{c^2}{2ac+2bc-c^2}\)

đặt pt là P

\(P\ge\frac{\left(a+b+c\right)^2}{2ab+2ac-a^2+2ab+2bc-b^2+2ac+2bc-c^2}\)

\(P\ge\frac{\left(a+b+c\right)^2}{4ab+4ac+4bc-a^2-b^2-c^2}\)

\(a^2+b^2+c^2\ge2ab+2bc+2ca\)(BĐT tương đương)

\(P\ge\frac{\left(a+b+c\right)^2}{4ab+4ac+4bc-a^2-b^2-c^2}\ge\frac{\left(a+b+c\right)^2}{2ab+2ac+2bc}\)

\(\left(a+b+c\right)^2\ge2ab+2ac+2bc\)(BĐT tương đương)

\(P\ge1\)

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