It's Mehul's birthday

Let $a=\sqrt x$ , $b=\sqrt y$ and $c=\sqrt z$ . Hence $abc=\sqrt {\ln{4.771}} = 1.25$

$a^2+b^2+c^2-2(ab+bc+ca)=4^2 ---- (1)$ $(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2+a^2)=a^4+b^4+c^4=40.125 ---- (2)$

Let $ab+bc+ca=k$ . Note that $a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2abc (a+b+c)=k^2-10$ Thus from equation (1) and (2), $(16-k)^2-2(k^2-10)=40.125$ $\Rightarrow 2k^2+64k+235.875=0$ $k=\dfrac {-64\pm 47}{4}$

Since $16-k=a^2+b^2+c^2>0$ , we have $x+y+z=a^2+b^2+c^2=16-k=\boxed{7.5}$