Don't try x = y = z

Rajen Sir's solution is probably the best but i solved this by only algebra,

$Let\quad a\& b\quad are\quad such\quad that\quad a+b=1\\ \\ Now\quad AM\ge GM\quad \quad \\ \\ \Rightarrow \cfrac { { x }^{ 2 }+ay^{ 2 } }{ 2 } \ge xy\sqrt { a } \quad \\ \\ \& \quad \cfrac { { by }^{ 2 }+z^{ 2 } }{ 2 } \ge yz\sqrt { b } \\ \\ if\quad \sqrt { \cfrac { a }{ b } } =\cfrac { \sqrt { 6 } }{ 4 } \Rightarrow a=\cfrac { 3 }{ 11 } ,b=\cfrac { 8 }{ 11 } \\ \\ \Rightarrow xy\sqrt { a } +yz\sqrt { b } \le \cfrac { { x }^{ 2 }+ay^{ 2 } }{ 2 } +\cfrac { { by }^{ 2 }+z^{ 2 } }{ 2 } \\ \\ \Rightarrow xy\sqrt { \cfrac { 3 }{ 11 } } +yz\sqrt { \cfrac { 8 }{ 11 } } \le \cfrac { 1 }{ 2 } \\ \\ \Rightarrow \sqrt { 6 } xy+4yz\le \cfrac { \sqrt { 22 } }{ 2 } \\ \\ \Rightarrow A=\cfrac { \sqrt { 22 } }{ 2 } \quad or\quad { A }^{ 2 }=\cfrac { 11 }{ 2 } =5.5$

Let $y = \cos\theta$ , $x = \sin\theta\cos\phi$ and $z = \sin\theta sin\phi$ and use the fact that $A\sin\phi + B \cos\phi$ has a maximum value $\sqrt{A^{2} + B^{2}}$ Putting in $\sqrt6 xy + 4yz = \sqrt2 y(\sqrt3 x + 2\sqrt2 z)$ and taking common $\sin\theta$ outside the bracket, and using the above mentioned maximizing $A = \sqrt{22}/2$ .

Let $\sqrt{6}xy + 4yz \leq k[x^{2} + y^{2} + z^{2}]$ for a real number $k$ . This way we just simply need to find the value of $k$ , which obviously is the maximum value of the above expression.

From AM-GM we have $k_{1} = \sqrt{6}xy \leq \frac{a_{1}}{2}x^{2} + \frac{a_{2}}{2}y^{2}$ for some real $a_{1}, a_{2}$ . With the same way we got $k_{2} = 4yz \leq \frac{a_{3}}{2}y^{2} + \frac{a_{1}}{2}z^{2}$

Therefore, $k_{1} + k_{2} \leq \frac{a_{1}}{2} [x^{2} + z^{2}] + [\frac{a_{2} + a_{3}}{2}]y^{2}$ We know that $a_{1}a_{2} = 6$ and $a_{3}a_{1} = 16$ While we want to find $a_{1}$ with $a_{2} + a_{3} = a_{1}$

Therefore, $\frac{6}{a_{1}} + \frac{16}{a_{1}} = a_{1}$ We can easily get $a_{1} = \sqrt{22}$ . Then we have $k_{1} + k_{2} \leq \frac{\sqrt{22}}{2}[x^{2} + y^{2} + z^{2}] = \frac{\sqrt{22}}{2} = A$ . Therefore $\boxed{A^{2} = \frac{11}{2}}$ .

Here is another solution,