Giải

$E=2{{x}^{2}}+8xy+11{{y}^{2}}-4x-2y+6$

$E=(2{{x}^{2}}+8{{y}^{2}}+2+8xy-4x-8y)+(3{{y}^{2}}+6y+3)+1$

$E=2({{x}^{2}}+4{{y}^{2}}+1+4xy-2x-4y)+3({{y}^{2}}+2y+1)+1$

$E=2{{(x+2y-1)}^{2}}+3{{(y+1)}^{2}}+1$

Ta có : ${{(x+2y-1)}^{2}}\ge 0;\,\,{{(y+1)}^{2}}\ge 0\,\,\forall x,y$

Do đó: $E\ge 1$

Dấu “=” xảy ra $\Leftrightarrow \left\{ \begin{align}& x+2y-1=0 \\ & y+1=0 \\
\end{align} \right.$$\Leftrightarrow \left\{ \begin{align}& x=3 \\ & y=-1 \\ \end{align} \right.$


Vậy GTNN của E = 1 khi $(x;y)=(3;-1)$


$F=2{{x}^{2}}+6{{y}^{2}}+5{{z}^{2}}-6xy+8yz-2xz+2y+4z+2$

$F=({{x}^{2}}+4{{y}^{2}}+{{z}^{2}}-4xy+4yz-2xz)+({{x}^{2}}-2xy+{{y}^{2}})+({{y}^{2}}+4{{z}^{2}}+1+4yz+2y+4z)+1$

$F={{(x-2y-z)}^{2}}+{{(x-y)}^{2}}+{{(y+2z+1)}^{2}}+1$

Vì ${{(x-2y-z)}^{2}}\ge 0;\,\,{{(x-y)}^{2}}\ge 0;\,\,{{(y+2z+1)}^{2}}\ge 0\,\,\forall x,y,z$ nên $F\ge 1$

Dấu “=” xảy ra $\Leftrightarrow \left\{ \begin{align}& x-2y-z=0 \\ & x-y=0 \\ & y+2z+1=0 \\
\end{align} \right.$$\Leftrightarrow \left\{ \begin{align}& x=1 \\ & y=1 \\ & z=-1 \\ \end{align} \right.$

Vậy GTNN của F = 1 khi $(x;y;z)=(1;1;-1)$