Triplication

Let us define as $F$ the midpoint of $\overline{BC}$ . $\overline{AE} = 2$ , $\overline{BE} = 4$ , $\overline{BF} = \overline{CF} = 2.5$ . By cosine law:

$\large \displaystyle \overline{BC}^2 = \overline{AC}^2 + \overline{AB}^2 -2 \cdot \overline{AC} \cdot \overline{AB}\ cos(\hat{A})$

$\large \displaystyle 25 = 16 + 36 - 48 cos(\hat{A})$

$\color{#20A900} \boxed{\large \displaystyle cos(\hat{A}) = \frac{9}{16} }$

$\large \displaystyle \overline{AC}^2 = \overline{AB}^2 + \overline{BC}^2 -2 \cdot \overline{AB}\cdot \overline{BC}\ cos(\hat{B})$

$\large \displaystyle 16 = 25 + 36 - 60 cos(\hat{B})$

$\color{#20A900} \boxed{\large \displaystyle cos(\hat{B}) = \frac{3}{4} }$

$\large \displaystyle \overline{CE}^2 = \overline{AC}^2 + \overline{AE}^2 -2 \cdot \overline{AC}\cdot \overline{AE}\ cos(\hat{A})$

$\large \displaystyle \overline{CE}^2 = 16 + 4 - 9$

$\color{#20A900} \boxed{\large \displaystyle \overline{CE} = \sqrt{11} }$

$\large \displaystyle \overline{EF}^2 = \overline{BE}^2 + \overline{BF}^2 -2 \cdot \overline{BE}\cdot \overline{BF}\ cos(\hat{B})$

$\large \displaystyle \overline{EF}^2 = \frac{25}{4} + 16 - 15$

$\color{#20A900} \boxed{\large \displaystyle \overline{EF} = \frac{\sqrt{29}}{2} }$

$\large \displaystyle \overline{CE} + \overline{EF} = \sqrt{11} + \frac{\sqrt{29}}{2}$

$\large \displaystyle \overline{CE} + \overline{EF} = \sqrt{ \left ( \sqrt{11} + \frac{\sqrt{29}}{2} \right )^2 }$

$\large \displaystyle \overline{CE} + \overline{EF} = \sqrt{ 11 + \frac{29}{4} + \sqrt{319} }$

$\large \displaystyle \overline{CE} + \overline{EF} = \frac12 \sqrt{ 73 + 4\sqrt{319} }$

$\color{#3D99F6} \boxed{\large \displaystyle A = 73} , \boxed{\large \displaystyle B = 4}, \boxed{\large \displaystyle C = 319}$

Basically same as Mr. Guilherme Niedu and MR. Marta Reece but little different.
$AE=2,~and~EB=4,~let~\angle CAB=\angle CAE=\alpha,~and~M~mdpoint~of~CB.\\ Using ~Cos~Rule~ in~\Delta s,~CAB~and~CAE,\\ Cos\alpha=\dfrac{4^2+6^2-5^2}{2*4*6}=9.\\ Also~ Cos\alpha=\dfrac{4^2+2^2-CE^2}{2*4*2}=20 - CE^2.\\ \therefore~CE=\sqrt{11}.\\ In ~\Delta~CEB,~median~EM=\frac 1 2 *\sqrt{\frac 1 2 *(CE^2+6^2 - 5^2)}=\frac 1 2\sqrt{29}.\\ \therefore~CE+EM=\frac 1 2 \sqrt{44}+\frac 1 2\sqrt{29}=\frac 1 2 \sqrt{44+29+Q\sqrt R}\\ \implies~P=\Large \color{#D61F06}{73} .$ .

From $\triangle ABC$ we get $cos(CAB)=\frac{4^2+6^2-5^2}{2\times4\times6}=\frac{9}{16}$

From $\triangle AEC$ and again the law of cosines $x=\sqrt{4^2+2^2-2\times2\times4\times cos(CAB)}=\sqrt{11}$

Again from $\triangle ABC$ comes $cos(ABC)=\frac{5^2+6^2-4^2}{2\times5\times6}=\frac{3}{4}$

And from $\triangle DEB$ we calculate $y=\sqrt{2.5^2+4^2-2\times2.5\times4\times cos(ABC)}=\frac{\sqrt{29}}{2}$

So $x+y=\sqrt{11}+\frac{\sqrt{29}}{2}$ .

This is a perfectly good answer. It does not, however, fit the required format. To do that, we have to show what it is a squire root of, in other words square it and put that under a square root. With a little simplification, that becomes.

$x+y=\frac{1}{2}\sqrt{73+4\sqrt{319}}$ .