Pack balls with maximum sum of radii We pack $8$ balls into a cube of side length $1$ so that no two balls share an interior point. what is the maximum sum of the radii of the balls?
It is possible to pack $8$ balls or radius $1/4$, which gives a sum of $2$, and it shouldn't be possible to do better. When we make balls unequal, we somehow lose some space due to the asymmetry.
 A: Your conjecture is wrong: $8$ balls of radius ${1\over4}$ are not optimal.
The following figure deals with the analogous problem in $2$ dimensions. One easily verifies (without computations) that the sum of the four radii is $>1$.

You can imitate this in dimension $3$. Put $4$ balls of radius $a$ into four corners of the cube, so that their centers $p_i$ form a regular tetrahedron. Their coordinates are
$$p_0=(a,a,a), \quad p_1=(a,1-a,1-a), \quad p_2=(1-a,a,1-a),\quad p_3=(1-a,1-a,a)\ .$$ Solving $|p_0-p_1|^2=(2a)^2$ for $a$ leads to $$a:=1-{1\over\sqrt{2}}\doteq0.2929\ .$$
Now we place  four spheres of radius $b$ into the other four corners of the cube. Their centers $q_i$ are
$$q_0=(1-b,1-b,1-b),\quad q_1=(1-b,b,b),\quad q_2=(b,1-b,b),\quad q_3=(b,b,1-b)\ .$$
The radius $b$ of the small spheres is chosen such that each small sphere  touches three larger spheres. Solving $|q_0-p_1|^2=(a+b)^2$ for $b$ then gives
$$b={1\over2}\left(3-\sqrt{2}-\sqrt{7-4\sqrt{2}}\>\right)\doteq0.213422\ .$$ The sum of the eight radii is 
$$4(a+b)\doteq2.02526>2\ .$$
