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All functions are smooth and continuous.
I use ' to signfy $d/dt$.
Given:

q_i'(0)=0, for i=1,2,3,4.
$q_2(t)q_3(t)>q_1(t)q_4(t)$ for $0<=t. T is finite. At t=T the inequality breaks.
q_4(0)*q_3''(0)>q_4''(0)*q_3(0)
(q_4''(0)+q_2''(0))(q_1(0)+q_3(0)>(q_4(0)+q_2(0))(q_1''(0)+q_3''(0))

This also holds but I suspect it is superfluous:
q_2(0)*q_1''(0)
One can also add this without loss of generality:
$q_1(0)=1, q_2(0)=-1$

Prove or disprove that:

q_4(t)*q_3''(t)>q_4''(t)*q_3(t)
(q_4''(t)+q_2''(t))(q_1(t)+q_3(t)>(q_4(t)+q_2(t))(q_1''(t)+q_3''(t))
for $0<=t.

Counterexample will do.

1 Answers 1

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It can't be true as you have nothing except smoothness to restrict the functions away from zero. Note that you are trying to prove (q_1(t)q_2'(t)-q_1'(t)q_2(t))'\gt 0 So start them off with $q_1(t)=a+bt^2, q_2(t)=c+dt^2, q_3(t)=e+ft^2, q_4(t)=g+ht^2$ for $t\le 1$. Your second equation can be satisfied if $c,e \gt \gt a,g$ Then for $t\gt 1$ add something smooth with positive derivative to $q_1$ like $\exp(\frac{a}{x^2-1})$ If you pick $a$ large enough, the derivatives will be larger than the function.