Let us start by drawing a picture. We are trying to find a lower bound on the angle between (OverVector[m]) and (OverVector[n]) (for q between 0 and 1 and χ between 0 and π/2) which is the same as the red angle depicted below.

Manipulate[
Module[{X,c,t,n,δ,ϵ,T,z,k},
X={(Sqrt[q+(q^2)])Sin[χ],q Cos[χ]};
If[χ==0,c={-((Sqrt[-q (-16-4 q+4 (q^2)+(q^3))])/(2 (1+q))),((2+(q^2))/(2+2 q))},c={(Csc[χ] (16 q+19 (q^2)+7 (q^3)+4 (q^4)-4 q (4+5 q+2 (q^2)+(q^3)) Cos[2 χ]-4 (q^2) Cos[3 χ]-4 (q^3) Cos[3 χ]+(q^2) Cos[4 χ]+(q^3) Cos[4 χ]-2 (Sqrt[2]) √(-(q^2) (1+q) (-32-45 q+8 (q^2)+8 (q^3)+8 (-16+q+4 (q^2)) Cos[χ]+(32+12 q-8 (q^2)) Cos[2 χ]-8 q Cos[3 χ]+q Cos[4 χ]) ((Sin[χ])^2))+Cos[χ] (4 (q^2)+4 (q^3)-2 (Sqrt[2]) q √(-(q^2) (1+q) (-32-45 q+8 (q^2)+8 (q^3)+8 (-16+q+4 (q^2)) Cos[χ]+(32+12 q-8 (q^2)) Cos[2 χ]-8 q Cos[3 χ]+q Cos[4 χ]) ((Sin[χ])^2)))))/(8 (Sqrt[q (1+q)]) (2+q+2 (q^2)+4 q Cos[χ]-q Cos[2 χ])),(8-2 q+8 q Cos[χ]+(q^2) Cos[χ]+4 (q^3) Cos[χ]+2 q Cos[2 χ]+4 (q^2) Cos[2 χ]-(q^2) Cos[3 χ]+(Sqrt[2]) √(-(q^2) (1+q) (-32-45 q+8 (q^2)+8 (q^3)+8 (-16+q+4 (q^2)) Cos[χ]+(32+12 q-8 (q^2)) Cos[2 χ]-8 q Cos[3 χ]+q Cos[4 χ]) ((Sin[χ])^2)))/(4 (2+q+2 (q^2)+4 q Cos[χ]-q Cos[2 χ]))}];
t=Normalize[{(Sqrt[q+(q^2)])Cos[χ],-q Sin[χ]}];
n=Normalize[{q Sin[χ],(Sqrt[q+(q^2)])Cos[χ]}];
{δ,ϵ}={ArcTan@@(c-X),ArcTan@@(-n)};
If[δ<0,δ=δ+2π];
If[ϵ<0,ϵ=ϵ+2π];
T={(1/(2 a ((a^2)+(((1+b))^2))))((a^4)+(a^2) (2 b+(b^2)+(((1+q))^2))+(1+b)a√(-((a^4)+4 (b^3)+(b^4)-4 b (((1+q))^2)+(((1+q))^2) (-3+2 q+(q^2))-2 (b^2) (-1+2 q+(q^2))+2 (a^2) (2 b+(b^2)-(((1+q))^2))))),(1/(2 ((a^2)+(((1+b))^2))))(-1-(a^2)-b+(a^2) b+(b^2)+(b^3)+2 q+2 b q+(q^2)+b (q^2)-a√(-((a^4)+4 (b^3)+(b^4)-4 b (((1+q))^2)+(((1+q))^2) (-3+2 q+(q^2))-2 (b^2) (-1+2 q+(q^2))+2 (a^2) (2 b+(b^2)-(((1+q))^2)))))}/.{a->c[[1]],b->c[[2]]};
{z,k}={ArcTan@@(c-T),ArcTan@@({0,-1}-T)};
If[z<0,z=z+2π];
If[k<0,k=k+2π];
Graphics[{
{Gray,Circle[{0,-1},1],Circle[c,1]},
{Blue,Circle[{0,0},{(Sqrt[q+(q^2)]),q}],AbsolutePointSize[5],Point[{0,0}],Text[Style["S",Italic,Bold],{0,-0.1}]},
{Red,Line[{X-2n,X,X+2(c-X)}],Circle[X,0.3,{δ,ϵ}],Line[{{2.3,1},{3.3,1},{3.3,1}+{-Cos[ϵ-δ],Sin[ϵ-δ]}}],AbsolutePointSize[5],Point[X],Text[Style["X",Italic,Bold],X+{0.05,0.08}],Circle[{3.3,1},0.3,{π-(ϵ-δ),π}],Dotted,Line[{X-t,X+t}]},
{RGBColor[0.4,0,1],Arrow[{X,X+n}],Text[Style["(OverVector[n])",Italic,Bold],X+(n/2)],Arrow[{X,2X-c}],Text[Style["(OverVector[m])",Italic,Bold],((3X-c)/2)]},
{Text[ϵ-δ,{2.8,0.8}]}
},AspectRatio->Automatic,PlotRange->{{-2,3.5},{-2,2}},ImageSize->Large]
],
{{q,1},0,1,Appearance->"Open"},{{χ,1.4},0,(π/2),Appearance->"Open"}]

We compute the center of the τ-ball which has X (determined by q and χ) on its boundary and which touches the drawn τ-ball, associated to S.

Simplify[Solve[(x^2)+(((y+1))^2)==4∧(((x-(Sqrt[q+(q^2)])Sin[χ]))^2)+(((y-q Cos[χ]))^2)==1,{x,y}],0<q<1∧0≤χ≤(π/2)]

The intersection we want is the one with the larger x-coordinate. We store this value into the variable center. From this we calculate the cotangent of both rays of the red angle. Using the addition theorem for cotangent, we obtain the cotangent of the red angle itself.

Module[{center,X,n,CX,ct1,ct2},
center={-((q ((Sqrt[2]) (Sqrt[-(1+q) (-32-45 q+8 (q^2)+8 (q^3)+8 (-16+q+4 (q^2)) Cos[χ]+(32+12 q-8 (q^2)) Cos[2 χ]-8 q Cos[3 χ]+q Cos[4 χ])])+(Sqrt[2]) q Cos[χ] (Sqrt[-(1+q) (-32-45 q+8 (q^2)+8 (q^3)+8 (-16+q+4 (q^2)) Cos[χ]+(32+12 q-8 (q^2)) Cos[2 χ]-8 q Cos[3 χ]+q Cos[4 χ])])-16 Sin[χ]-19 q Sin[χ]-7 (q^2) Sin[χ]-4 (q^3) Sin[χ]-4 q Sin[2 χ]-4 (q^2) Sin[2 χ]+q Sin[3 χ]+(q^2) Sin[3 χ]))/(4 (Sqrt[q (1+q)]) (2+q+2 (q^2)+4 q Cos[χ]-q Cos[2 χ]))),(1/(4 (2+q+2 (q^2)+4 q Cos[χ]-q Cos[2 χ])))(8-2 q+8 q Cos[χ]+(q^2) Cos[χ]+4 (q^3) Cos[χ]+2 q Cos[2 χ]+4 (q^2) Cos[2 χ]-(q^2) Cos[3 χ]+(Sqrt[2]) q (Sqrt[-(1+q) (-32-45 q+8 (q^2)+8 (q^3)+8 (-16+q+4 (q^2)) Cos[χ]+(32+12 q-8 (q^2)) Cos[2 χ]-8 q Cos[3 χ]+q Cos[4 χ])]) Sin[χ])};
X={(Sqrt[q+(q^2)])Sin[χ],q Cos[χ]};
n={q Sin[χ],(Sqrt[q+(q^2)])Cos[χ]};
CX=center-X;
ct1=((CX[[1]])/(CX[[2]]));
ct2=((n[[1]])/(n[[2]]));
Simplify[((ct1 ct2+1)/(ct1-ct2)),0<q<1∧0≤χ≤(π/2)]
]

We define the function ct, which measures the cotangent of the red angle.

ct[q_,χ_]:=(-2 (-8-7 q+13 (q^2)+12 (q^3)) Cos[χ]+q (8+4 q-12 (q^2)-8 (q^3)-8 Cos[2 χ]-4 q Cos[2 χ]+4 (q^2) Cos[2 χ]+2 Cos[3 χ]+2 q Cos[3 χ]-2 (Sqrt[2]) (Sqrt[-(1+q) (-32-45 q+8 (q^2)+8 (q^3)+8 (-16+q+4 (q^2)) Cos[χ]+(32+12 q-8 (q^2)) Cos[2 χ]-8 q Cos[3 χ]+q Cos[4 χ])]) Sin[χ]+(Sqrt[2]) (Sqrt[-(1+q) (-32-45 q+8 (q^2)+8 (q^3)+8 (-16+q+4 (q^2)) Cos[χ]+(32+12 q-8 (q^2)) Cos[2 χ]-8 q Cos[3 χ]+q Cos[4 χ])]) Sin[2 χ]))/((Sqrt[q (1+q)]) (-2 (Sqrt[2]) q (Sqrt[-(1+q) (-32-45 q+8 (q^2)+8 (q^3)+8 (-16+q+4 (q^2)) Cos[χ]+(32+12 q-8 (q^2)) Cos[2 χ]-8 q Cos[3 χ]+q Cos[4 χ])])-2 (Sqrt[2]) Cos[χ] (Sqrt[-(1+q) (-32-45 q+8 (q^2)+8 (q^3)+8 (-16+q+4 (q^2)) Cos[χ]+(32+12 q-8 (q^2)) Cos[2 χ]-8 q Cos[3 χ]+q Cos[4 χ])])-16 Sin[χ]+2 q Sin[χ]+8 (q^2) Sin[χ]+8 Sin[2 χ]+6 q Sin[2 χ]-4 (q^2) Sin[2 χ]-6 q Sin[3 χ]+q Sin[4 χ]));

We claim that ct is increasing in q for every χ.

Manipulate[
Plot[ct[q,χ],{q,0,1},AspectRatio->Automatic,PlotRange->{{0,1},{-1,1}}],
{{χ,1.2},0,(π/2),Appearance->"Open"}]

To verify this, we prove that the partial derivative of ct with respect to q, which we denote by dct, is ≥0.

Simplify[Together[D[ct[q,χ],q]],0<q<1∧0≤χ≤(π/2)]

dct[q_,χ_]:=(4 (2+q+2 (q^2)+4 q Cos[χ]-q Cos[2 χ]) (16 (Sqrt[2])+258 (Sqrt[2]) q+586 (Sqrt[2]) (q^2)+424 (Sqrt[2]) (q^3)+144 (Sqrt[2]) (q^4)+64 (Sqrt[2]) (q^5)+192 (Sqrt[2]) Cos[χ]+464 (Sqrt[2]) q Cos[χ]+448 (Sqrt[2]) (q^2) Cos[χ]+368 (Sqrt[2]) (q^3) Cos[χ]+192 (Sqrt[2]) (q^4) Cos[χ]+127 (Sqrt[2]) q Cos[2 χ]+175 (Sqrt[2]) (q^2) Cos[2 χ]+32 (Sqrt[2]) (q^3) Cos[2 χ]-16 (Sqrt[2]) (q^4) Cos[2 χ]+64 (Sqrt[2]) Cos[3 χ]+56 (Sqrt[2]) q Cos[3 χ]+8 (Sqrt[2]) (q^2) Cos[3 χ]+16 (Sqrt[2]) (q^3) Cos[3 χ]-16 (Sqrt[2]) Cos[4 χ]-2 (Sqrt[2]) q Cos[4 χ]+6 (Sqrt[2]) (q^2) Cos[4 χ]-8 (Sqrt[2]) (q^3) Cos[4 χ]-8 (Sqrt[2]) q Cos[5 χ]-8 (Sqrt[2]) (q^2) Cos[5 χ]+(Sqrt[2]) q Cos[6 χ]+(Sqrt[2]) (q^2) Cos[6 χ]-4 (Sqrt[-(1+q) (-32-45 q+8 (q^2)+8 (q^3)+8 (-16+q+4 (q^2)) Cos[χ]+(32+12 q-8 (q^2)) Cos[2 χ]-8 q Cos[3 χ]+q Cos[4 χ])]) Sin[χ]-16 q (Sqrt[-(1+q) (-32-45 q+8 (q^2)+8 (q^3)+8 (-16+q+4 (q^2)) Cos[χ]+(32+12 q-8 (q^2)) Cos[2 χ]-8 q Cos[3 χ]+q Cos[4 χ])]) Sin[χ]-32 (q^2) (Sqrt[-(1+q) (-32-45 q+8 (q^2)+8 (q^3)+8 (-16+q+4 (q^2)) Cos[χ]+(32+12 q-8 (q^2)) Cos[2 χ]-8 q Cos[3 χ]+q Cos[4 χ])]) Sin[χ]+8 (Sqrt[-(1+q) (-32-45 q+8 (q^2)+8 (q^3)+8 (-16+q+4 (q^2)) Cos[χ]+(32+12 q-8 (q^2)) Cos[2 χ]-8 q Cos[3 χ]+q Cos[4 χ])]) Sin[2 χ]+8 q (Sqrt[-(1+q) (-32-45 q+8 (q^2)+8 (q^3)+8 (-16+q+4 (q^2)) Cos[χ]+(32+12 q-8 (q^2)) Cos[2 χ]-8 q Cos[3 χ]+q Cos[4 χ])]) Sin[2 χ]+16 (q^2) (Sqrt[-(1+q) (-32-45 q+8 (q^2)+8 (q^3)+8 (-16+q+4 (q^2)) Cos[χ]+(32+12 q-8 (q^2)) Cos[2 χ]-8 q Cos[3 χ]+q Cos[4 χ])]) Sin[2 χ]-4 (Sqrt[-(1+q) (-32-45 q+8 (q^2)+8 (q^3)+8 (-16+q+4 (q^2)) Cos[χ]+(32+12 q-8 (q^2)) Cos[2 χ]-8 q Cos[3 χ]+q Cos[4 χ])]) Sin[3 χ]))/(q (1+q) (Sqrt[-q (-32-45 q+8 (q^2)+8 (q^3)+8 (-16+q+4 (q^2)) Cos[χ]+(32+12 q-8 (q^2)) Cos[2 χ]-8 q Cos[3 χ]+q Cos[4 χ])]) (((-2 (Sqrt[2]) q (Sqrt[-(1+q) (-32-45 q+8 (q^2)+8 (q^3)+8 (-16+q+4 (q^2)) Cos[χ]+(32+12 q-8 (q^2)) Cos[2 χ]-8 q Cos[3 χ]+q Cos[4 χ])])-2 (Sqrt[2]) Cos[χ] (Sqrt[-(1+q) (-32-45 q+8 (q^2)+8 (q^3)+8 (-16+q+4 (q^2)) Cos[χ]+(32+12 q-8 (q^2)) Cos[2 χ]-8 q Cos[3 χ]+q Cos[4 χ])])-16 Sin[χ]+2 q Sin[χ]+8 (q^2) Sin[χ]+8 Sin[2 χ]+6 q Sin[2 χ]-4 (q^2) Sin[2 χ]-6 q Sin[3 χ]+q Sin[4 χ]))^2));

Plot3D[{0,dct[q,χ]},{q,0,1},{χ,0,(π/2)},ClippingStyle->None]

Manipulate[
Plot[dct[q,χ],{q,0,1},AspectRatio->Automatic,PlotRange->{{0,1},{0,3}}],
{{χ,1.2},0,(π/2),Appearance->"Open"}]

Since ct is increasing in the first variable, it follows that any upper bound of ct[1, _ ] will also be an upper bound for the whole ct.

Simplify[ct[1,χ],0≤χ≤(π/2)]

Plot[ct[1,χ],{χ,0,π/2},AspectRatio->Automatic,PlotRange->{{0,π/2},{0,1}}]

This is also an increasing function; let us verify that its derivative is positive.

Simplify[D[(Sec[(χ/2)] (4+10 Cos[χ]+4 Cos[2 χ]-2 Cos[3 χ]+2 (Sqrt[61+88 Cos[χ]-36 Cos[2 χ]+8 Cos[3 χ]-Cos[4 χ]]) Sin[χ]-(Sqrt[61+88 Cos[χ]-36 Cos[2 χ]+8 Cos[3 χ]-Cos[4 χ]]) Sin[2 χ]))/(4 (Sqrt[2]) (Cos[(χ/2)] (Sqrt[61+88 Cos[χ]-36 Cos[2 χ]+8 Cos[3 χ]-Cos[4 χ]])+2 (1-4 Cos[χ]+Cos[2 χ]) ((Sin[(χ/2)])^3))),χ],0≤χ≤(π/2)]

Plot[-((2 (Sqrt[2]) (-5 Cos[(χ/2)]+Cos[((3 χ)/2)]) ((Sin[(χ/2)])^2) (20 Cos[(χ/2)] (Sqrt[61+88 Cos[χ]-36 Cos[2 χ]+8 Cos[3 χ]-Cos[4 χ]])+4 Cos[((3 χ)/2)] (Sqrt[61+88 Cos[χ]-36 Cos[2 χ]+8 Cos[3 χ]-Cos[4 χ]])-102 Sin[(χ/2)]-12 Sin[((3 χ)/2)]-12 Sin[((5 χ)/2)]-5 Sin[((7 χ)/2)]+Sin[((9 χ)/2)]))/((Sqrt[61+88 Cos[χ]-36 Cos[2 χ]+8 Cos[3 χ]-Cos[4 χ]]) (((Cos[(χ/2)] (Sqrt[61+88 Cos[χ]-36 Cos[2 χ]+8 Cos[3 χ]-Cos[4 χ]])+2 (1-4 Cos[χ]+Cos[2 χ]) ((Sin[(χ/2)])^3)))^2))),{χ,0,π/2},AspectRatio->Automatic,PlotRange->{{0,π/2},{0,1}}]

The denominator is clearly positive. Let us examine the numerator.

Numerator[-((2 (Sqrt[2]) (-5 Cos[(χ/2)]+Cos[((3 χ)/2)]) ((Sin[(χ/2)])^2) (20 Cos[(χ/2)] (Sqrt[61+88 Cos[χ]-36 Cos[2 χ]+8 Cos[3 χ]-Cos[4 χ]])+4 Cos[((3 χ)/2)] (Sqrt[61+88 Cos[χ]-36 Cos[2 χ]+8 Cos[3 χ]-Cos[4 χ]])-102 Sin[(χ/2)]-12 Sin[((3 χ)/2)]-12 Sin[((5 χ)/2)]-5 Sin[((7 χ)/2)]+Sin[((9 χ)/2)]))/((Sqrt[61+88 Cos[χ]-36 Cos[2 χ]+8 Cos[3 χ]-Cos[4 χ]]) (((Cos[(χ/2)] (Sqrt[61+88 Cos[χ]-36 Cos[2 χ]+8 Cos[3 χ]-Cos[4 χ]])+2 (1-4 Cos[χ]+Cos[2 χ]) ((Sin[(χ/2)])^3)))^2)))]

The first part of the numerator is positive, as the following shows.

-2 (Sqrt[2]) (-5 Cos[(χ/2)]+Cos[((3 χ)/2)]) ((Sin[(χ/2)])^2) 
==
2 (Sqrt[2]) (5 Cos[(χ/2)]-(4((Cos[(χ/2)])^3)-3Cos[(χ/2)]))((Sin[(χ/2)])^2) 
==
8(Sqrt[2]) Cos[(χ/2)](2-((Cos[(χ/2)])^2))((Sin[(χ/2)])^2)

We can simplify the final factor as follows.

20 Cos[(χ/2)] (Sqrt[61+88 Cos[χ]-36 Cos[2 χ]+8 Cos[3 χ]-Cos[4 χ]])
+4 Cos[((3 χ)/2)] (Sqrt[61+88 Cos[χ]-36 Cos[2 χ]+8 Cos[3 χ]-Cos[4 χ]])
-102 Sin[(χ/2)]-12 Sin[((3 χ)/2)]-12 Sin[((5 χ)/2)]-5 Sin[((7 χ)/2)]+Sin[((9 χ)/2)]
==
(20 Cos[(χ/2)]+4(4((Cos[(χ/2)])^3)-3Cos[(χ/2)]))(Sqrt[61+88 Cos[χ]-36 Cos[2 χ]+8 Cos[3 χ]-Cos[4 χ]])
-102 Sin[(χ/2)]-12 Sin[((3 χ)/2)]-12 Sin[((5 χ)/2)]-5 Sin[((7 χ)/2)]+Sin[((9 χ)/2)]
==
8Cos[(χ/2)](1+2((Cos[(χ/2)])^2))(Sqrt[61+88 Cos[χ]-36 Cos[2 χ]+8 Cos[3 χ]-Cos[4 χ]])
-102 Sin[(χ/2)]-12 Sin[((3 χ)/2)]-12 Sin[((5 χ)/2)]-5 Sin[((7 χ)/2)]+Sin[((9 χ)/2)]

The first line is clearly positive. If the second line is also positive, the total is positive. Otherwise, the positivity of this expression is equivalent to the positivity of the difference of the square of the first line and the square of the second line.

Simplify[(((8Cos[(χ/2)](1+2((Cos[(χ/2)])^2))(Sqrt[61+88 Cos[χ]-36 Cos[2 χ]+8 Cos[3 χ]-Cos[4 χ]])))^2)-(((-102 Sin[(χ/2)]-12 Sin[((3 χ)/2)]-12 Sin[((5 χ)/2)]-5 Sin[((7 χ)/2)]+Sin[((9 χ)/2)]))^2),0≤χ≤(π/2)]

This is positive, as the following shows.

18241+37815 Cos[χ]+11096 Cos[2 χ]+1290 Cos[3 χ]+580 Cos[4 χ]+66 Cos[5 χ]+40 Cos[6 χ]-(7/2) Cos[7 χ]-5 Cos[8 χ]+(1/2) Cos[9 χ]
≥
18241-11096-1290-580-66-40-(7/2)-5-(1/2)

18241-11096-1290-580-66-40-(7/2)-5-(1/2)

Hence the maximum of ct is

ct[1,(π/2)]

Since the red angle (the cotangent of which is ct) is between 0 and π, it follows that it is ≥ ArcCot[(1/(Sqrt[2]))]. This is therefore also true for the angle between (OverVector[m]) and (OverVector[n]) which is what we are interested in.

The angle (Subscript[α, S])X is 0 if pr(X)-X points inside the ellipsoid (Subscript[E, p])(S), otherwise it is complementary to the angle between (OverVector[m]) and (OverVector[n]). Either way 0≤(Subscript[α, S])X≤ArcTan[(1/(Sqrt[2]))]. Equivalently, 0≤(Subscript[α, S])X≤ArcCos[(Sqrt[(2/3)])]; in particular Cos[(Subscript[α, S])X]≥(Sqrt[(2/3)]).

Cos[ArcTan[(1/(Sqrt[2]))]]