I really hate when I see a post with you do that and that and rest is easy but in fact, the rest is what makes the problem hard. This is such post, but really, the rest of it is as not so important as the first part. Maybe I had made a mistake in calculation somewhere, but problem can be solved like this for sure. Put the origin of Cartesian system in $A$ such that $A(0,0,0),B(4,0,0),C(4,4,0)...$ (we put $a=1$). Then it's not hard to se that $P(2,y,4)$ and $Q(x,0,\frac{x+2}2)$ where $x,y\in [0,4]$. So $CP+PQ=\sqrt{20+(y-4)^{2}}+\sqrt{(x-2)^{2}+(\frac{x+2}2-4)^{2}+y^{2}}$. We want to calculate the minimum of this expression when $x,y\in [0,4]$. The minimum of $(x-2)^{2}+(\frac{x+2}2-4)^{2}$ can easily be calculated, and if I am not wrong it is $\frac65$ and it is obtained for $x=\frac{16}5\in [0,4]$ (you see that only this part depends on $x$, so we can calculate its minimum and use it in further solving). Now we should calculate minimum of $\sqrt{(y-4)^{2}+20}+\sqrt{y^{2}+\frac{6}5}$. If you put $M(4,\sqrt{20})$ and $N(0,\sqrt{\frac{6}{5}})$, then $\sqrt{(y-4)^{2}+20}+\sqrt{y^{2}+\frac{6}5}$ is $YM+YN$ where $Y(y,0)$. This is well known problem, and solution is (again if I am not wrong, i really calculated all this in maximum speed) is $y=\frac{4\sqrt{6}}{10+4\sqrt{5}}$. This is all for $a=1$, but only multiply everything whit $a$, and that's it.

I didn't even think about solving this with coordinates! I guess you made some mistaken calculations... The answer is $\sqrt{\frac{276}{5}}a$. I took it "easier". I expressed $CP+PQ$ by the right triangles $CHP$ and $QGP$. Setting $HP=x$, we get $CP+PQ=\sqrt{CH^{2}+x^{2}}+\sqrt{QG^{2}+(4a-x)^{2}}$ and we use Minkovski. So, to attain the minimum, $QG$ must be firstly minimal, that is $Q$ is the projection of $G$ on $EF$. Then we apply Minkovski to "get rid" of $x$. So minimum is $\sqrt{(CH+QG)^{2}+(4a)^{2}}$. Well, I don't make all the calculations. It is easy to get $CH$ from right triangle $CHP$, and $QG$ as being altitude in $EGF$.

Well, I like coordinate geometry, and this is the first time I use it on 3D that's why I was so exciting, and didn't look for other "easier" solutions . I usually use it in 2D like in http://www.mathlinks.ro/Forum/viewtopic.php?t=136680v Bye

freemind wrote: I didn't even think about solving this with coordinates! I guess you made some mistaken calculations... The answer is $\sqrt{\frac{276}{5}}a$. I took it "easier". I expressed $CP+PQ$ by the right triangles $CHP$ and $QGP$. Setting $HP=x$, we get $CP+PQ=\sqrt{CH^{2}+x^{2}}+\sqrt{QG^{2}+(4a-x)^{2}}$ and we use Minkovski. So, to attain the minimum, $QG$ must be firstly minimal, that is $Q$ is the projection of $G$ on $EF$. Then we apply Minkovski to "get rid" of $x$. So minimum is $\sqrt{(CH+QG)^{2}+(4a)^{2}}$. Well, I don't make all the calculations. It is easy to get $CH$ from right triangle $CHP$, and $QG$ as being altitude in $EGF$. I think apply Minkowski is wrong because your equation never attain $\sqrt{\frac{276}{5}}a$. I think to solve your equation need derivative equalize to zero and $x$ will be nearly $2.38$ and answer is nearly $7.47938$