Bump....

Bump$~~~~~$

Very elegant, here is a solution. Sorry for poor English Let the vertices of the tetrahedron be $V_1,V_2,V_3,V_4,$ $M_1,M_2,M_3,M_4$ are midpoint of $V_1V_4,V_1V_3,V_2V_3,V_2V_4.$ Call the surface $\alpha =M_1M_2M_3M_4,$ the idea is to find a plane perpendicular to $\alpha.$ [asy][asy] unitsize(2cm); pair V[]; pair V1,V2,V3,V4; pair M[]; pair M1,M2,M3,M4; V[1]=(0,0); V1=V[1]; V[2]=(-1,-4);V2=V[2]; V[3]=(1.5,-3.8);V3=V[3]; V[4]=(3,-2.5);V4=V[4]; M[1]=midpoint(V[1]--V[4]); M1=M[1]; M2=midpoint(V1--V3); M3=midpoint(V2--V3); M4=midpoint(V2--V4); path p1=V1--V2--V3--V4--cycle; path p2=M1--M2--M3--M4--cycle; filldraw(p2,red+white,blue+dashed+.75); draw(V2--V4,dashed); draw(p1^^V1--V3); label("$V_1$",V1, dir(V3--V1)); label("$V_2$",V2,-dir(V2--V1,V2--V3)); label("$V_3$",V3, dir(V1--V3)); label("$V_4$",V4,-dir(V4--V3,V4--V1)); label("$M_1$",M1,0.75*dir(V4--V1)*dir(-90)); label("$M_2$",M2, dir(M2--V1,M2--M3)); label("$M_3$",M3,dir(V3--V2)*dir(90)); label("$M_4$",M4,dir(M4--V3,M4--V4)); //label("$a$",V1--V2);// AoPS asy autosizes label("$a$",midpoint(V1--V2), dir(V2--V1)*dir(90)); label("$\alpha$", midpoint(M2--M3),dir(M2--M3)*dir(90)); //label("$b$", V3--V4);// AoPS asy autosizes label("$b$", midpoint(V4--V3), dir(V4--V3)*dir(90)); real gr=(sqrt(5)-1)/2; dot(p1^^p2,black+gr,Fill(red+white)); shipout(bbox(2mm,Fill(white))); [/asy][/asy] In this way, the projection turns out to be a trapezoid, and the height is a constant. Therefore we only need to calculate the sum of the projection length of $a=V_1V_2$ and $b=V_3V_4.$ Now the problem is only on 2-dimension space: For two given vectors $v_1,v_2,$ there exists two lines $\ell_1,\ell_2$ such that the ratio of sum of projection length is no less than $\sqrt 2.$ WLOG let $|v_1|\ge |v_2|,$ then let $\ell _1$ be perpendicular to $v_1$ so the sum is no bigger than $|v_2|.$ Let $\ell_2$ parellel to $v_1\pm v_2,$ we can get the sum is no less than $|v_1\pm v_2|.$ Now it is obvious because $|v_1+v_2|^2+|v_1-v_2|^2=2(|v_1|^2+|v_2|^2)\ge 4|v_2|^2,$ and we are done.$\Box$