It is as tough as it looks

Note: I previously Posted Similiar Solution some time ago So here I only copy Paste it :) "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 : $AC = 5$ . ${ E }_{ min }\quad =\quad { A }_{ min }\quad =\quad 5\quad \quad \quad$ . "

For Calculating Value of 'x' Let's Redraw the diagram , Now 'D' must lies on AC for to minimizes The given expression (Condition for Triangle equality )

$Area(\Delta ADB)+Area(\Delta CDB)\quad =\quad Area(\Delta ABC)\\ \cfrac { 1 }{ 2 } (4x)\sin { (45) } \quad +\quad \cfrac { 1 }{ 2 } (3x)\sin { (45) } =\cfrac { 1 }{ 2 } (4)(3)\sin { (90) } \\ \boxed { x\quad =\quad \cfrac { 12\sqrt { 2 } }{ 7 } }$ .

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

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

$\quad \quad \space = \dfrac {x - \frac {3}{\sqrt{2}}} {\sqrt{(x-\frac {3} {\sqrt{2}})^2 + \frac {9}{2}} } + \dfrac {x - 2\sqrt{2}} {\sqrt{(x-2\sqrt{2})^2 + 8}} = 0$

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

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

$\Rightarrow \left( \dfrac {\sqrt{(x-\frac {3} {\sqrt{2}})^2 + \frac {9}{2}} } {x - \frac {3}{\sqrt{2}}} \right)^2 = \left(- \dfrac {\sqrt{(x-2\sqrt{2})^2 + 8}} {x - 2\sqrt{2}} \right)^2$

$\Rightarrow 1 + \dfrac {\frac {9}{2}} {\left( x - \frac {3}{\sqrt{2}}\right)^2 } = 1 + \dfrac {8} {\left( x - 2\sqrt{2} \right)^2}$

$\Rightarrow \dfrac {\frac {9}{2}} {\left( x - \frac {3}{\sqrt{2}}\right)^2 } = \dfrac {8} {\left( x - 2\sqrt{2} \right)^2}$

$\Rightarrow 9 \left( x - 2\sqrt{2} \right)^2 = 16 \left( x - \frac {3}{\sqrt{2}}\right)^2 \quad \Rightarrow 3 \left( x - 2\sqrt{2} \right) = \pm 4 \left( x - \frac {3}{\sqrt{2}}\right)$

$\Rightarrow \begin {cases} 3 \left( x - 2\sqrt{2} \right) = 4 \left( x - \frac {3}{\sqrt{2}}\right) &\Rightarrow x = 0 \\ 3 \left( x - 2\sqrt{2} \right) = - 4 \left( x - \frac {3}{\sqrt{2}}\right) & \Rightarrow x = \dfrac {12\sqrt{2}} {7} \end {cases}$

$\Rightarrow a = 12$ , $b=2$ and $c=7\quad \Rightarrow a+b+c = 12+2+7 = \boxed{21}$

Given that first differential is zero, A being constant be geometrically a straight line formed by the third sides of two triangles, both having 45deg angles between 3 and x and between x and 4, as cosine formula if applied forms the given expression in x . Cumulatively, angle between 3 and 4 is 90deg , giving A = 5 hypotenuse. Now x is the length of angle bisector of right - angle of this right-angled triangle to be calculated using sine formula for each of the triangles to give a = 12, b = 2 and c = 7.

Consider points P:(x,0), A:(3/√2,3/√2) and B;(2√2,2√2). Now PA + PB =√(x²-3x√2+9)+√(x²-4x√2+16). Now consider a reflection of point A in the x-axis at A': (3/√2, -3/√2). Wherever BA' intersects with the x-axis, at that x, (PA+PB) or the given expression will be minimum giving us the necessary value of x as 12√2/7.

we can also consider two vectors, the given expression is addition of magnitude of two vectors which is always greater than or equal to magnitude of their resultant. using this,we find value of x for which the two vectors have are collinear, that is the answer