xem lại đề; x = 1 -> đề sai

Đề bài có lẽ bị sai , nếu thử x = 5 , y = 7 , z = 8 

Đặt \(f\left(x\right)=x^3+a^2x^2-ax-6\)

Áp dụng định lý Bezout:

\(f\left(x\right)=x^3+a^2x^2-ax-6⋮x-1\Leftrightarrow f\left(1\right)=0\)

\(\Leftrightarrow1+a^2-a-6=0\Leftrightarrow a^2-a-5=0\)

\(\Leftrightarrow a^2-a=5\Leftrightarrow a^2-a+\frac{1}{4}=\frac{21}{4}\)

\(\Leftrightarrow\left(a-\frac{1}{2}\right)^2=\frac{21}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}a=\frac{\sqrt{21}}{2}+\frac{1}{2}\\a=\frac{-\sqrt{21}}{2}+\frac{1}{2}\end{cases}}\)

Vậy \(\orbr{\begin{cases}a=\frac{\sqrt{21}}{2}+\frac{1}{2}\\a=\frac{-\sqrt{21}}{2}+\frac{1}{2}\end{cases}}\)thì \(x^3+a^2x^2-ax-6⋮x-1\)

= x⁵-x⁴+x³-x³-1=x²(x²-x+1) -(x+1)(x^2 -x+1)=(x²-x-1)(x²-x-1)

= x⁵ -x⁴+x³-x³-1=x³(x²-x+1)-(x+1)(x^2-x+1)=(x²-x+1)(x³-x-1)

Ta chứng minh \(x^4+y^4\ge x^3y+xy^3\)

\(\Leftrightarrow x^3\left(x-y\right)-y^3\left(x-y\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\left[\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}\right]\ge0\)(luôn đúng)

Áp dụng vào bài toán ta có:

\(x^4+y^4\ge x^3y+xy^3\)\(\Rightarrow2\left(x^4+y^4\right)\ge x^4+y^4+x^3y+xy^3\)\(=\left(x^3+y^3\right)\left(x+y\right)\)

\(\Rightarrow\frac{x^4+y^4}{x^3+y^3}\ge\frac{x+y}{2}\).Tương tự ta cũng có:

\(\frac{y^4+z^4}{y^3+z^3}\ge\frac{y+z}{2};\frac{z^4+x^4}{z^3+x^3}\ge\frac{z+x}{2}\)

Cộng theo vế ta có: \(VT\ge\frac{x+y}{2}+\frac{y+z}{2}+\frac{z+x}{2}=x+y+z=1\)

Dấu = khi \(x=y=z=\frac{2008}{3}\)