@Lia - Maths is fun !

\(Let:a,b,c\ge0\text{ }such:a+b+c=3.Found\text{ }max\text{ }and\text{ }min\text{ }A=\sqrt{x+3}+\sqrt{y+3}+\sqrt{z+3}\)    

My solution !

*Found max

Using Bunhiacopxki we have

\(A^2\le\left(a+3+b+3+c+3\right)\left(1+1+1\right)=...=36\)

\(\Rightarrow A\le6\left(Because\:\text{ }\text{ }A\ge0\text{ }so\text{ }A\text{ }can't\text{ }< 0\text{ }\right)\)

\(A_{max}=6\text{ }\Leftrightarrow a=b=c=1\)

*Found min

We have extra inequality \(\sqrt{x+z}+\sqrt{y+z}\ge\sqrt{z}+\sqrt{x+y+z}\left(x;y;z\ge0\right)\)(1)

Prove : \(\left(1\right)\Leftrightarrow x+y+2z+2\sqrt{\left(x+z\right)\left(y+z\right)}\ge z+x+y+z+2\sqrt{z\left(x+y+z\right)}\)

                     \(\Leftrightarrow\sqrt{xy+xz+yz+z^2}\ge\sqrt{xz+yz+z^2}\)        

                    \(\Leftrightarrow xy+xz+yz+z^2\ge xz+yz+z^2\)

                    \(\Leftrightarrow xy\ge0\left(True!\right)\)

Using (1) we have

\(A=\sqrt{a+3}+\sqrt{b+3}+\sqrt{c+3}\ge\sqrt{3}+\sqrt{a+b+3}+\sqrt{c+3}\)

                                                                                 \(=\sqrt{3}+\sqrt{3}+\sqrt{a+b+c}\)

                                                                                  \(=3\sqrt{3}\)

\(A_{min}=3\sqrt{3}\text{ }when\text{ }\hept{\begin{cases}a=b=\frac{3}{2}\\c=0\end{cases}}\)

       (In here I using when because there are many other a,b,c such a = 0 ; b = c = ³/₂)

The problem is done !