Với mọi x;y;z ta luôn có:
\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx\ge0\)
\(\Leftrightarrow x^2+y^2+z^2\ge xy+yz+zx\)
\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)\ge3\left(xy+yz+zx\right)\)
\(\Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)
\(\Rightarrow xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}=\frac{3^2}{3}=3\)
\(B_{max}=3\) khi \(x=y=z=1\)
\(f\left(x+1\right)=\left(x+1\right)^2+p\left(x+1\right)+q\)
\(=x^2+2x+1+px+p+q\)
\(=\left(x^2+px+q\right)+2x+q+1=f\left(x\right)+2x+p+1\)
\(\Rightarrow f\left(x\right).f\left(x+1\right)=f\left(x\right)\left[f\left(x\right)+2x+p+1\right]\)
\(=f^2\left(x\right)+2x.f\left(x\right)+x^2-x^2+p.f\left(x\right)+f\left(x\right)\)
\(=\left[f\left(x\right)+x\right]^2-x^2+p.f\left(x\right)+x^2+px+q\)
\(=\left[f\left(x\right)+x\right]^2+p\left[f\left(x\right)+x\right]+q\)
\(=f\left[f\left(x\right)+x\right]\)
Thay \(x=2008\Rightarrow f\left(2008\right).f\left(2009\right)=f\left[f\left(2008\right)+2008\right]\)
Vậy tồn tại số nguyên k sao cho \(f=f\left(2008\right)+2008\) để \(f\left(k\right)=f\left(2008\right).f\left(2009\right)\)