b. ta có:

\(x^2+y^2+z^2-\left(2xy-2zx+2yz\right)\)

\(=x^2+y^2+z^2-2xy-2zx+2yz\)

\(=\left(x-y+z\right)^2\ge0\)

do đó \(x^2+y^2+z^2\ge2xy-2xz+2yz\)

a.ta có:

\(x^2+y^2+z^2-\left(xy+yz+zx\right)\)

\(=\frac{1}{2}\left[\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)\right]\)

\(=\frac{1}{2}\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\ge0\)

vì \(\left(x-y\right)^2\ge0,\left(y-z\right)^2\ge0,\left(z-x\right)^2\ge0\)

do đó : 

\(x^2+y^2+z^2\ge xy+yz+zx\)

dấu = xảy ra khi và chỉ khi x-y-z

b. \(x^3+3x^2-4\)  =\(x^3-x^2+4x^2-4\)

                             =\(x^2\left(x-1\right)+4\left(x-1\right)\left(x^2+4x+4\right)\)

                             =\(\left(x-1\right)\left(x+2\right)^2\)

a+1