Not very difficult

$A = \sqrt{9+x^2-3x\sqrt{2}} + \sqrt{16+x^2-4x\sqrt{2}}$

$\quad = \sqrt{x^2-\frac {6}{ \sqrt{2}}x +\frac {9}{2} +\frac {9}{2}} + \sqrt{ x^2-4x\sqrt{2} + 8 + 8}$

$\quad = \sqrt{(x-\frac {3}{ \sqrt{2}})^2 +\frac {9}{2}} + \sqrt{ (x-2\sqrt{2} )^2 + 8 }$

$\quad = \sqrt{(x-\frac {3}{ \sqrt{2}})^2 + (0 -\frac {3}{ \sqrt{2}})^2} + \sqrt{ (x-2\sqrt{2} )^2 +(0 -2\sqrt{2})^2 }$

Therefore, $A$ is the sum of lengths of two straight lines $(x,0)$ , $(\frac {3}{ \sqrt{2}},\frac {3}{ \sqrt{2}})$ and $(x,0)$ , $(2\sqrt{2}, 2\sqrt{2})$ .

The shortest total length is that traveled by light from $(\frac {3}{ \sqrt{2}},\frac {3}{ \sqrt{2}})$ to $(2\sqrt{2}, 2\sqrt{2})$ reflected at the $x$ -axis at $(x,0)$ . And the angle of incident = angle of reflection.

Therefore, $\dfrac {x - \frac {3}{ \sqrt{2}}} {\frac {3}{ \sqrt{2}}} = \dfrac {2\sqrt{2}-x}{ 2 \sqrt{2} } \quad \Rightarrow \frac {\sqrt{2}} {3} x - 1 = 1 - \frac {1} {2\sqrt{2}} x$

$\Rightarrow (\frac {\sqrt{2}} {3} + \frac {1} {2\sqrt{2}} ) x = 2 \quad \Rightarrow (\frac {\sqrt{2}} {3} + \frac {\sqrt{2}} {4}) x = 2 \quad \Rightarrow \frac {7\sqrt{2}} {12} x = 2 \quad \Rightarrow x = \frac {12\sqrt{2}} {7}$

Substitute $x = \frac {12\sqrt{2}} {7}$ in $A$ , we have:

$A_{min} = \sqrt{( \frac {12\sqrt{2}} {7} -\frac {3\sqrt{2}} {2})^2 +\frac {9}{2}} + \sqrt{ ( \frac {12\sqrt{2}} {7} -2\sqrt{2} )^2 + 8 }$

$\quad \quad = \sqrt{( \frac {3\sqrt{2}} {14} )^2 +\frac {9}{2}} + \sqrt{( -\frac {2\sqrt{2}} {7} )^2 + 8 }$

$\quad \quad = \sqrt{\frac {18} {196} +\frac {9}{2}} + \sqrt{\frac {8} {49} + 8 } = \sqrt{\frac {900} {196}} + \sqrt{\frac {400} {49}} = \boxed{5}$

Let's Do some geometric ! Let's The given expression in question be E

Now Inter-prate that This expression is written in form of cosine law .

According to figure : Figure

E = AD + DC

and By Triangle inequality in Triangle ADC :

$AD\quad +\quad DC\quad \ge \quad AC$ .

And For calculating AC note that Triangle ABC is right angled triangle.

So By Pythogoras Theoram :

$AC = 5$ .

${ E }_{ min }\quad =\quad { A }_{ min }\quad =\quad 5\quad \quad \quad$ .

Q.E.D

First Of all differentiate $'A'$

$\Large{\frac { d }{ dx } \left( \frac { \sqrt { 9+{ x }^{ 2 }-3x\sqrt { 2 } } }{ 1 } +\frac { \sqrt { 16+{ x }^{ 2 }-4x\sqrt { 2 } } }{ 1 } \right) \\ =\left( \frac { 2x-4\sqrt { 2 } }{ 2\sqrt { x^{ 2 }-4x\sqrt { 2 } +16 } } +\frac { 2x-3\sqrt { 2 } }{ 2\sqrt { 9+{ x }^{ 2 }-3x\sqrt { 2 } } } \right)}$

Now to maximize it

$\Large{\left( \frac { 2x-4\sqrt { 2 } }{ 2\sqrt { x^{ 2 }-4x\sqrt { 2 } +16 } } +\frac { 2x-3\sqrt { 2 } }{ 2\sqrt { 9+{ x }^{ 2 }-3x\sqrt { 2 } } } \right) =0\\ \therefore \quad x=\frac { 12\sqrt { 2 } }{ 7 }}$

Now placing value of $x$ in $'A'$ we get $\boxed{\large{A=5}}$

differentiate once,then equate to zero.....you'll get the value of "x",substitute in the equation and ....................job done :)