$x^2+y^2+z^2+t^2≥x(y+z+t)$
$=x^2+y^2+z^2+t^2≥xy+xz+xt$
$=4x^2+4y^2+4z^2+4t^2≥4xy+4xz+4xt$
$=(x-2y)^2+(x-2z)^2+(x-2t)^2+x^2≥0$
Vì: $\begin{cases}(x-2y)^2≥0∀x,y\\(x-2z)^2≥0∀x,z\\(x-2t)^2≥0∀x,t\\x^2≥0∀x\end{cases}$
$⇒(x-2y)^2+(x-2z)^2+(x-2t)^2+x^2≥0∀x,y,z$
Dấu "=" xảy ra $⇔x=y=z=t$