Find min of $IA + IB + IC +ID$ in tetrahedron $ABCD$ Let the point $I$ in tetrahedron $ABCD$. Find $\min\{IA + IB + IC + ID\}$.
I can't solve this problem, even in the case ABCD regular. Please help


I only can solve regular case, here is the solution:
$H$ and $F$ are one the plane $BCD$, $H,F$ is the foot of $A,I$,
$G$ is on $AH$ and $GH=IF=q,IG=FH=p,\angle FHB=\theta,BH=HD=HB=b,AH=h $
$AI=\sqrt{p^2+(h-q)^2},BF^2=p^2+b^2-2pbcos(\theta),FD^2=p^2+b^2-2pbcos(\dfrac{2\pi}{3}-\theta),FC^2=p^2+b^2-2pbcos(\dfrac{2\pi}{3}+\theta)$ 
$AI+BI+CI+DI=\sqrt{p^2+(h-q)^2}+\sqrt{p^2-2pbcos(\theta)+b^2+q^2}+\sqrt{p^2+b^2-2pbcos(\dfrac{2\pi}{3}-\theta)+q^2}+\sqrt{p^2+b^2-2pbcos(\dfrac{2\pi}{3}+\theta)+q^2}=f(p,q,\theta)$ 
we note that $p,q$ is independent, so we take $p$ first to let $f_{p}=f_{p_{min}}$  
$f'_{p}=\dfrac{2p}{\sqrt{p^2+(h-q)^2}}+\dfrac{2p-2bcos(\theta)}{\sqrt{p^2-2pbcos(\theta)+b^2+q^2}}+\dfrac{2p-2bcos(\dfrac{2\pi}{3}-\theta)}{\sqrt{p^2+b^2-2pbcos(\dfrac{2\pi}{3}-\theta)}}+\dfrac{2p-2bcos(\dfrac{2\pi}{3}+\theta)}{\sqrt{p^2+b^2-2pbcos(\dfrac{2\pi}{3}+\theta)+q^2}}=0$
here we proof that $g(x)=\dfrac{2x-2m}{\sqrt{x^2-2mx+m^2+q^2}}$ will be mono increasing function. because:
$g'(x)=\dfrac{q^2}{(x^2-2mx+m^2+q^2)^{\frac{3}{2}}}>0 \to f'_{p}$is mono increasing function also . luckily $f'_{p}(0)=0$, so we have only one root. it is trivial that $\sqrt{p^2-tp+s}$ will be increasing function when $p$ is big, so $ p=0 \to f_{p_{min}}$  and again we are lucky to clean $\theta$ also.
now $f=3\sqrt{b^2+q^2}+h-q$, $b,h$ is known so you can find final result.
