a=2009,b=2010,c=2011

M=4(2009-2010)(2010-2011)=(2009-2011)^2=4

Đặt \(\frac{a}{2009}=\frac{b}{2010}=\frac{c}{2011}=k\)

=>a=2009k;b=2010k;c=2011k

Xét \(4\left(a-b\right)\left(b-c\right)=4\left(2009k-2010k\right)\left(2010k-2011k\right)\)

\(=4\left(-k\right)\left(-k\right)=4k^2\left(1\right)\)

Xét \(\left(c-a\right)^2=\left(2011k-2009k\right)^2=\left(2k\right)^2=4k^2\left(2\right)\)

Từ (1) và (2)

=>4(a-b)(b-c)=(c-a)²=4k²

Hay M=4k²

\(a^2 + b^2 + c^2 = 4\)

=>\(\begin{cases}a^2\le4\\ b^2\le4\\ c^2\le4\end{cases}\)

=>-2<=a<=2 và -2<=b<=2 và -2<=c<=2

=>2-a>=0; 2-b>=0; 2-c>=0

2-a>=0

\(a^2\ge0\)

Do đó: \(a^2(2-a)\ge0\)

=>\(2a^2-a^3\ge0\)

=>\(a^3\le2a^2\)

Chứng minh tương tự, ta cũng có: \(b^3\le2b^2;c^3\le2c^2\)

Do đó: \(a^3+b^3+c^3\le2\left(a^2+b^2+c^2\right)=2\cdot4=8\)

Dấu '=' xảy ra khi \(\begin{cases}a^2\left(2-a\right)=0\\ b^2\left(2-b\right)=0\\ c^2\left(2-c\right)=0\end{cases}\)

=>a;b;c∈{0;2}

mà \(a^2+b^2+c^2=4\)

nên a=2; b=c=0

\(P = a^4 + b^4 + c^4 = 2^4 + 0^4 + 0^4 = 16\)

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

\(\Leftrightarrow14+2\left(ab+bc+ac\right)=0\Leftrightarrow ab+bc+ac=-7\)

Suy ra : \(\left(ab+bc+ac\right)^2=49\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=49\)

\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=49\)

\(a^2+b^2+c^2=14\Leftrightarrow\left(a^2+b^2+c^2\right)^2=196\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=196\)

\(\Leftrightarrow a^4+b^4+c^4+2.49=256\)  \(\Leftrightarrow a^4+b^4+c^4=98\)

Vậy ... 

\(a+b+c=0\)

\(\Rightarrow\left(a+b+c\right)^2=0\)

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

\(\Leftrightarrow2ab+2bc+2ca=-14\)

\(\Leftrightarrow ab+bc+ca=-7\)

\(\Rightarrow\left(ab+bc+ca\right)^2=49\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2ab^2c+2abc^2+2a^2bc=49\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=49\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=49\).

\(a^2+b^2+c^2=14\)

\(\Rightarrow\left(a^2+b^2+c^2\right)^2=14^2=196\)

\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=196\)

\(\Leftrightarrow a^4+b^4+c^4+2.49=196\)

\(\Leftrightarrow a^4+b^4+c^4=98\)