Lemma 5.6.

Collect[Expand[(1+t^2)(e0+e2 t^2+e4 t^4)-(c0+c2 t^2)^3-(e4-c2^3)(t^2-h1)(t^2-h2)(t^2-h3)],t]

Solve[{-(c0^3)+e0-(c2^3) h1 h2 h3+e4 h1 h2 h3==0,-3 (c0^2) c2+e0+e2+(c2^3) h1 h2-e4 h1 h2+(c2^3) h1 h3-e4 h1 h3+(c2^3) h2 h3-e4 h2 h3==0,-3 c0 (c2^2)+e2+e4-(c2^3) h1+e4 h1-(c2^3) h2+e4 h2-(c2^3) h3+e4 h3==0},{e0,e2,e4}]

e0:=-(-(c0^3)-(c0^3) h1-(c0^3) h2-(c0^3) h1 h2-(c0^3) h3-(c0^3) h1 h3-(c0^3) h2 h3-3 (c0^2) c2 h1 h2 h3+3 c0 (c2^2) h1 h2 h3-(c2^3) h1 h2 h3)/((1+h1) (1+h2) (1+h3))

e2:=-((c0^3)-3 (c0^2) c2+(c0^3) h1-3 (c0^2) c2 h1+(c0^3) h2-3 (c0^2) c2 h2-3 c0 (c2^2) h1 h2+(c2^3) h1 h2+(c0^3) h3-3 (c0^2) c2 h3-3 c0 (c2^2) h1 h3+(c2^3) h1 h3-3 c0 (c2^2) h2 h3+(c2^3) h2 h3-3 c0 (c2^2) h1 h2 h3+(c2^3) h1 h2 h3)/((1+h1) (1+h2) (1+h3))

e4:=-(-(c0^3)+3 (c0^2) c2-3 c0 (c2^2)-(c2^3) h1-(c2^3) h2-(c2^3) h1 h2-(c2^3) h3-(c2^3) h1 h3-(c2^3) h2 h3-(c2^3) h1 h2 h3)/((1+h1) (1+h2) (1+h3))

Collect[Expand[a+b z+k1 x^2+k2 y^2+k3 z^2+d z(x^2+y^2+z^2)+e(x^2+y^2+z^2)^2/.{x->x1,y->y1,z->z1+t}],{x1,y1,z1}]

Factor[d+4 e t+2 s (k3+3 d t+6 e t^2)+2 s^2(b+2 k3 t+3 d t^2+4 e t^3)/.{s->-(b+2 k3 t+3 d t^2+4 e t^3)/(4(b t+k3 t^2+d t^3+e t^4))}]

t-equation is:

Factor[-(-(b^3)-2 (b^2) k3 t-5 (b^2) d (t^2)-20 (b^2) e (t^3)+5 b (d^2) (t^4)-20 b e k3 (t^4)+4 b d e (t^5)+2 (d^2) k3 (t^5)-8 e (k3^2) (t^5)+(d^3) (t^6)+8 b (e^2) (t^6)-4 d e k3 (t^6))/.{b->-1,k1->c0,k2->c2,k3->(e0-e2+e4)/(c0-c2)^2,d->((c0+c2)e2-2 c2 e0-2 c0 e4)/(c0-c2)^2,e->(c2^2 e0-c0 c2 e2+c0^2 e4)/(c0-c2)^2}]

The solution t for the third factor is:

Solve[(1+h1+h2+h1 h2+h3+h1 h3+h2 h3+h1 h2 h3-2 c0 t-2 c0 h1 t-2 c0 h2 t-2 c0 h1 h2 t-2 c2 h3 t-2 c2 h1 h3 t-2 c2 h2 h3 t-2 c2 h1 h2 h3 t+(c0^2) (t^2)+(c0^2) h1 (t^2)+(c0^2) h2 (t^2)+2 c0 c2 h1 h2 (t^2)-(c2^2) h1 h2 (t^2)-(c0^2) h3 (t^2)+2 c0 c2 h3 (t^2)+(c2^2) h1 h3 (t^2)+(c2^2) h2 h3 (t^2)+(c2^2) h1 h2 h3 (t^2))==0,t]

denominator:

Simplify[(2 ((c0^2)+(c0^2) h1+(c0^2) h2+2 c0 c2 h1 h2-(c2^2) h1 h2-(c0^2) h3+2 c0 c2 h3+(c2^2) h1 h3+(c2^2) h2 h3+(c2^2) h1 h2 h3))]

square root part:

Simplify[((((-2 c0-2 c0 h1-2 c0 h2-2 c0 h1 h2-2 c2 h3-2 c2 h1 h3-2 c2 h2 h3-2 c2 h1 h2 h3))^2)-4 (1+h1+h2+h1 h2+h3+h1 h3+h2 h3+h1 h2 h3) ((c0^2)+(c0^2) h1+(c0^2) h2+2 c0 c2 h1 h2-(c2^2) h1 h2-(c0^2) h3+2 c0 c2 h3+(c2^2) h1 h3+(c2^2) h2 h3+(c2^2) h1 h2 h3))]

other part:

Simplify[2 c0+2 c0 h1+2 c0 h2+2 c0 h1 h2+2 c2 h3+2 c2 h1 h3+2 c2 h2 h3+2 c2 h1 h2 h3]

Therefore, setting w:c0/c2, k:(Sqrt[(1+h1) (1+h2) (h1-h3) (h2-h3)]) ,t:((w+h3)(1+h1)(1+h2)+(w-1)k)/(c2((1+h1+h2-h3)w^2+2(h1 h2+h3)w+h1 h3+h2 h3-h1 h2+h1 h2 h3)),

First we prove s:-(b+2 k3 t+3 d t^2+4 e t^3)/(4(b t+k3 t^2+d t^3+e t^4))

is equal to s:- c2 (-h1 h2+h1 h3+h2 h3+h1 h2 h3+(2 h1 h2+2 h3) w+(1+h1+h2-h3) (w^2))k/{2(1+h1)  (1+h2) (h1-h3) (h2-h3) (-1+w) }

We consider the difference (times the denominator) :

ds:=-(b+2 k3 t+3 d t^2+4 e t^3)-(4(b t+k3 t^2+d t^3+e t^4))(- c2 (-h1 h2+h1 h3+h2 h3+h1 h2 h3+(2 h1 h2+2 h3) w+(1+h1+h2-h3) (w^2))k/{2(1+h1)  (1+h2) (h1-h3) (h2-h3) (-1+w) })

Simplify[Factor[(ds/.{{b->-1,k1->c0,k2->c2,k3->(e0-e2+e4)/(c0-c2)^2,d->((c0+c2)e2-2 c2 e0-2 c0 e4)/(c0-c2)^2,e->(c2^2 e0-c0 c2 e2+c0^2 e4)/(c0-c2)^2}})/.{c0->w c2,t->((w+h3)(1+h1)(1+h2)+(w-1)k)/(c2((1+h1+h2-h3)w^2+2(h1 h2+h3)w+h1 h3+h2 h3-h1 h2+h1 h2 h3))}]/.{k->√ (1+h1)(1+h2) (h1-h3) (h2-h3)}]

Therefore we get  ds=0.

Here we introduce  u, v as follows:

u:=h1+2 (h1^2)+(h1^3)+h2-2 h1 h2-(h1^2) h2+2 (h1^3) h2+2 (h2^2)-h1 (h2^2)-4 (h1^2) (h2^2)+(h2^3)+2 h1 (h2^3)-2 h3+h1 h3+2 (h1^2) h3-(h1^3) h3+h2 h3+(h1^2) h2 h3+2 (h2^2) h3+h1 (h2^2) h3-(h2^3) h3-4 (h3^2)-h1 (h3^2)+2 (h1^2) (h3^2)-h2 (h3^2)-2 h1 h2 (h3^2)+2 (h2^2) (h3^2)-2 (h3^3)-h1 (h3^3)-h2 (h3^3)

v:=2-2 (h1^2)+4 h1 h2-2 (h2^2)+4 h3+2 (h3^2)

Next  we calculate  1/e'''=1/a'=1/(b t+k3 t^2+d t^3+e t^4)

We prove  1/e'''=-c2((u+v k) (w+(((h1 h2+h3+k)/(1+h1+h2-h3))))^4)/((w-1)^3(((h1-h2))^2)  (h1-h3) (h2-h3) (1+h3))

To do so, we consider the difference:

die3:=1-(b t+k3 t^2+d t^3+e t^4)(-c2((u+v k) (w+(((h1 h2+h3+k)/(1+h1+h2-h3))))^4)/((w-1)^3(((h1-h2))^2)  (h1-h3) (h2-h3) (1+h3)))

Simplify[((die3/.{{b->-1,k1->c0,k2->c2,k3->(e0-e2+e4)/(c0-c2)^2,d->((c0+c2)e2-2 c2 e0-2 c0 e4)/(c0-c2)^2,e->(c2^2 e0-c0 c2 e2+c0^2 e4)/(c0-c2)^2}})/.{c0->w c2,t->((w+h3)(1+h1)(1+h2)+(w-1)k)/(c2((1+h1+h2-h3)w^2+2(h1 h2+h3)w+h1 h3+h2 h3-h1 h2+h1 h2 h3))})/.{k->√ (1+h1)(1+h2) (h1-h3) (h2-h3)}]

Simplify[u+v√ (1+h1)(1+h2) (h1-h3) (h2-h3)/.{h3->r,h2->q+r,h1->p+q+r}]

Expand[(p(1+2 q+r)+2(1+q+(q^2)+2 r+q r+(r^2)))^2-4q (p+q) (1+q+r) (1+p+q+r)]

As for  k3'''=(k3+3d t+6e t^2)+3(b+2k3 t+3d t^2+4e t^3)s+6(b t+k3 t^2+d t^3+e t^4)s^2

we prove  k3'''=(c2 (h1+h2+2 h1 h2-2 h3-h1 h3-h2 h3) (-1+w))/(2 (1+h1) (1+h2) (1+h3))

Consider the difference

dk3:=(k3+3d t+6e t^2)+3(b+2k3 t+3d t^2+4e t^3)s+6(b t+k3 t^2+d t^3+e t^4)s^2 -(c2 (h1+h2+2 h1 h2-2 h3-h1 h3-h2 h3) (-1+w))/(2 (1+h1) (1+h2) (1+h3))

Simplify[((dk3/.{{b->-1,k1->c0,k2->c2,k3->(e0-e2+e4)/(c0-c2)^2,d->((c0+c2)e2-2 c2 e0-2 c0 e4)/(c0-c2)^2,e->(c2^2 e0-c0 c2 e2+c0^2 e4)/(c0-c2)^2}})/.{c0->w c2,t->((w+h3)(1+h1)(1+h2)+(w-1)k)/(c2((1+h1+h2-h3)w^2+2(h1 h2+h3)w+h1 h3+h2 h3-h1 h2+h1 h2 h3)),s->- c2 (-h1 h2+h1 h3+h2 h3+h1 h2 h3+(2 h1 h2+2 h3) w+(1+h1+h2-h3) (w^2))k/{2(1+h1)  (1+h2) (h1-h3) (h2-h3) (-1+w) }})/.{k->√ (1+h1)(1+h2) (h1-h3) (h2-h3)}]

As for k2''':=k2+d t+2e t^2+(b+2k3 t+3d t^2+4e t^3)s+2(b t+k3 t^2+d t^3+e t^4)s^2

We prove  k2'''=-((c2 (2+h1+h2) (-1+w))/(2 (1+h1) (1+h2)))

Consider the difference

dk2:=k2+d t+2e t^2+(b+2k3 t+3d t^2+4e t^3)s+2(b t+k3 t^2+d t^3+e t^4)s^2-(-((c2 (2+h1+h2) (-1+w))/(2 (1+h1) (1+h2))))

Simplify[((dk2/.{{b->-1,k1->c0,k2->c2,k3->(e0-e2+e4)/(c0-c2)^2,d->((c0+c2)e2-2 c2 e0-2 c0 e4)/(c0-c2)^2,e->(c2^2 e0-c0 c2 e2+c0^2 e4)/(c0-c2)^2}})/.{c0->w c2,t->((w+h3)(1+h1)(1+h2)+(w-1)k)/(c2((1+h1+h2-h3)w^2+2(h1 h2+h3)w+h1 h3+h2 h3-h1 h2+h1 h2 h3)),s->- c2 (-h1 h2+h1 h3+h2 h3+h1 h2 h3+(2 h1 h2+2 h3) w+(1+h1+h2-h3) (w^2))k/{2(1+h1)  (1+h2) (h1-h3) (h2-h3) (-1+w) }})/.{k->√ (1+h1)(1+h2) (h1-h3) (h2-h3)}]

As for  k1''':=k1+d t+2e t^2+(b+2k3 t+3d t^2+4e t^3)s+2(b t+k3 t^2+d t^3+e t^4)s^2

We prove k1'''=((c2 (h1+h2+2 h1 h2) (-1+w))/(2 (1+h1) (1+h2)))

Consider the difference

dk1:=k1+d t+2e t^2+(b+2k3 t+3d t^2+4e t^3)s+2(b t+k3 t^2+d t^3+e t^4)s^2-(((c2 (h1+h2+2 h1 h2) (-1+w))/(2 (1+h1) (1+h2))))

Simplify[((dk1/.{{b->-1,k1->c0,k2->c2,k3->(e0-e2+e4)/(c0-c2)^2,d->((c0+c2)e2-2 c2 e0-2 c0 e4)/(c0-c2)^2,e->(c2^2 e0-c0 c2 e2+c0^2 e4)/(c0-c2)^2}})/.{c0->w c2,t->((w+h3)(1+h1)(1+h2)+(w-1)k)/(c2((1+h1+h2-h3)w^2+2(h1 h2+h3)w+h1 h3+h2 h3-h1 h2+h1 h2 h3)),s->- c2 (-h1 h2+h1 h3+h2 h3+h1 h2 h3+(2 h1 h2+2 h3) w+(1+h1+h2-h3) (w^2))k/{2(1+h1)  (1+h2) (h1-h3) (h2-h3) (-1+w) }})/.{k->√ (1+h1)(1+h2) (h1-h3) (h2-h3)}]

As for  a''':=e+(d+4e t)s+(k3+3d t+6e t^2)s^2+(b+2k3 t+3d t^2+4e t^3)s^3+(b t+k3 t^2+d t^3+e t^4)s^4

We prove  a'''=-c2^3(u+v k)(w+(((h1 h2+h3+k)/(1+h1+h2-h3))))^4/(16 (((1+h1))^2) (((1+h2))^2) (h1-h3) (h2-h3) (1+h3) (-1+w))

Consider the difference

da:=e+(d+4e t)s+(k3+3d t+6e t^2)s^2+(b+2k3 t+3d t^2+4e t^3)s^3+(b t+k3 t^2+d t^3+e t^4)s^4-(-c2^3(u+v k)(w+(((h1 h2+h3+k)/(1+h1+h2-h3))))^4/(16 (((1+h1))^2) (((1+h2))^2) (h1-h3) (h2-h3) (1+h3) (-1+w)))

Simplify[((da/.{{b->-1,k1->c0,k2->c2,k3->(e0-e2+e4)/(c0-c2)^2,d->((c0+c2)e2-2 c2 e0-2 c0 e4)/(c0-c2)^2,e->(c2^2 e0-c0 c2 e2+c0^2 e4)/(c0-c2)^2}})/.{c0->w c2,t->((w+h3)(1+h1)(1+h2)+(w-1)k)/(c2((1+h1+h2-h3)w^2+2(h1 h2+h3)w+h1 h3+h2 h3-h1 h2+h1 h2 h3)),s->- c2 (-h1 h2+h1 h3+h2 h3+h1 h2 h3+(2 h1 h2+2 h3) w+(1+h1+h2-h3) (w^2))k/{2(1+h1)  (1+h2) (h1-h3) (h2-h3) (-1+w) }})/.{k->√ (1+h1)(1+h2) (h1-h3) (h2-h3)}]

Therefore A:=-k1'''/(2 e''') is (under u->U, v->V)

Simplify[(-((c2 (h1+h2+2 h1 h2) (-1+w))/(2 (1+h1) (1+h2))))(-c2((U+V k) (w+(((h1 h2+h3+k)/(1+h1+h2-h3))))^4)/((w-1)^3(((h1-h2))^2)  (h1-h3) (h2-h3) (1+h3)))/2]

A0:=((c2^2) (h1+h2+2 h1 h2) (U+k V) (((((h1 h2+h3+k)/(1+h1+h2-h3))+w))^4))/(4 (1+h1) (((h1-h2))^2) (1+h2) (h1-h3) (h2-h3) (1+h3) (((-1+w))^2))

B:=-k2'''/(2 e''') is

Simplify[-(-((c2 (2+h1+h2) (-1+w))/(2 (1+h1) (1+h2))))(-c2((U+V k) (w+(((h1 h2+h3+k)/(1+h1+h2-h3))))^4)/((w-1)^3(((h1-h2))^2)  (h1-h3) (h2-h3) (1+h3)))/2]

B0:=-((c2^2) (2+h1+h2) (U+k V) (((((h1 h2+h3+k)/(1+h1+h2-h3))+w))^4))/(4 (1+h1) (((h1-h2))^2) (1+h2) (h1-h3) (h2-h3) (1+h3) (((-1+w))^2))

C:=-k3'''/(2 e''') is

Simplify[-((c2 (h1+h2+2 h1 h2-2 h3-h1 h3-h2 h3) (-1+w))/(2 (1+h1) (1+h2) (1+h3)))(-c2((U+V k) (w+(((h1 h2+h3+k)/(1+h1+h2-h3))))^4)/((w-1)^3(((h1-h2))^2)  (h1-h3) (h2-h3) (1+h3)))/2]

C0:=((c2^2) (h1+h2+2 h1 h2-2 h3-h1 h3-h2 h3) (U+k V) (((((h1 h2+h3+k)/(1+h1+h2-h3))+w))^4))/(4 (1+h1) (((h1-h2))^2) (1+h2) (h1-h3) (h2-h3) (((1+h3))^2) (((-1+w))^2))

D^2:=a'''/e'''  is

Simplify[(-c2^3(U+V k)(w+(((h1 h2+h3+k)/(1+h1+h2-h3))))^4/(16 (((1+h1))^2) (((1+h2))^2) (h1-h3) (h2-h3) (1+h3) (-1+w)))(-c2((U+V k) (w+(((h1 h2+h3+k)/(1+h1+h2-h3))))^4)/((w-1)^3(((h1-h2))^2)  (h1-h3) (h2-h3) (1+h3)))]

Therefore D is

((c2^2) (U+k V) (((((h1 h2+h3+k)/(1+h1+h2-h3))+w))^4))/(4 (1+h1) (h1-h2) (1+h2) (h1-h3) (h2-h3) (1+h3) (((-1+w))^2))

D0:=((c2^2) (U+k V) (((((h1 h2+h3+k)/(1+h1+h2-h3))+w))^4))/(4 (1+h1) (h1-h2) (1+h2) (h1-h3) (h2-h3) (1+h3) (((-1+w))^2))

Simplify[B0/A0]

Simplify[C0/A0]

Simplify[D0/A0]

Since A0>0, we have B0<0,D0>0. Further

Simplify[(A0-C0)/A0]

Simplify[(C0-D0)/A0]

Simplify[(-B0-D0)/A0]

Hence we have  A0>C0>D0>0, -B0>D0.

Further

Simplify[(A0-C0)/(C0-B0)]

Simplify[(A0-D0)/(D0-B0)]

Simplify[(A0+D0)/(-B0-D0)]