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Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
$A\geq \frac{9}{x+2+y+2+z+2}=\frac{9}{x+y+z+6}$
Áp dụng BĐT Bunhiacopxky:
$(x^2+y^2+z^2)(1+1+1)\geq (x+y+z)^2$
$\Rightarrow 9\geq (x+y+z)^2\Rightarrow x+y+z\leq 3$
$\Rightarrow A\geq \frac{9}{x+y+z+6}\geq \frac{9}{3+6}=1$
Vậy $A_{\min}=1$. Dấu "=" xảy ra khi $x=y=z=1$
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\(2\left(x+y\right)+xy=x^2+y^2\\ \Leftrightarrow x^2+y^2-2x-2y-xy=0\\ \Leftrightarrow2x^2+2y^2-4x-4y-2xy=0\\ \Leftrightarrow\left(x^2-4x+4\right)+\left(y^2-4y+4\right)+\left(x^2-2xy+y^2\right)=8\\ \Leftrightarrow\left(x-2\right)^2+\left(y-2\right)^2+\left(x-y\right)^2=8\)
\(\Leftrightarrow\begin{matrix}\left(x-2\right)^2=0;&\left(y-2\right)^2=4;&\left(x-y\right)^2=4\\\left(x-2\right)^2=4;&\left(y-2\right)^2=0;&\left(x-y\right)^2=4\\\left(x-2\right)^2=4;&\left(y-2\right)^2=4;&\left(x-y\right)^2=0\end{matrix}\)
\(\Leftrightarrow\begin{matrix}x=2;&y=4\\x=2;&y=0\\x=4;&y=2\\x=0;&y=2\\x=0;&y=0\\x=2;&y=2\end{matrix}\)
Vậy có 6 cặp số thỏa mãn:
\(\left(x;y\right)\in\left\{\left(2;4\right);\left(2;0\right);\left(4;2\right);\left(0;2\right);\left(0;0\right);\left(2;2\right)\right\}\)
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\(A\le\sqrt{3\left(x+y+y+z+z+x\right)}=\sqrt{6\left(x+y+z\right)}\le\sqrt{6.\sqrt{3\left(x^2+y^2+z^2\right)}}=\sqrt{6\sqrt{3}}\)
\(A_{max}=\sqrt{6\sqrt{3}}\) khi \(x=y=z=\dfrac{1}{\sqrt{3}}\)
Do \(x^2+y^2+z^2=1\Rightarrow0\le x;y;z\le1\)
\(\Rightarrow\left\{{}\begin{matrix}x^2\le x\\y^2\le y\\z^2\le z\end{matrix}\right.\) \(\Rightarrow x+y+z\ge x^2+y^2+z^2=1\)
\(A^2=2\left(x+y+z\right)+2\sqrt{\left(x+y\right)\left(x+z\right)}+2\sqrt{\left(x+y\right)\left(y+z\right)}+2\sqrt{\left(y+z\right)\left(z+x\right)}\)
\(A^2=2\left(x+y+z\right)+2\sqrt{x^2+xy+yz+zx}+2\sqrt{y^2+xy+yz+zx}+2\sqrt{z^2+xy+yz+zx}\)
\(A^2\ge2\left(x+y+z\right)+2\sqrt{x^2}+2\sqrt{y^2}+2\sqrt{z^2}=4\left(x+y+z\right)\ge4\)
\(\Rightarrow A\ge2\)
\(A_{min}=2\) khi \(\left(x;y;z\right)=\left(0;0;1\right)\) và các hoán vị
\(\left(y+2\right)x^2+1=y^2\)
\(\left(y+2\right)x^2-3=\left(y-2\right)\left(y+2\right)\)
\(\left(y+2\right)\left(x^2-y+2\right)=3\)
vì x,y e Z nên y+2 và x2-y+2 đều thuộc Z
nên chúng là ước của 3
còn lại tự bấm máy