I seek at least one instance of positive fractions $p,q,r$ where $p,q,r \in \mathbb{Q}^+$, for which $$\frac12 (a - 2 b + c)^2 \ge pa^2 + qb^2 + rc^2\quad \forall\ a,b,c \in \mathbb{R}.$$ I used this code

FindInstance[{1/2 (a - 2 b + c)^2 >= 
    p*a^2 + q*b^2 + r*c^2 && {p, q, r} ∈ 
    Rationals && {a, b, c} ∈ Reals }, {p, q, r}]

But it simply returns the code to me, with an error that the system contains independent variables $a,b,c$.

Next, I replaced FindInstance with Reduce and I got many solutions with conditions on $a,b,c$. I would like to restrict them to the values of $p,q,r$ that work $\forall\ a,b,c$. What is the correct way to do this?


The following shows that Reduce thinks there is no answer:

        {a,b,c} ∈ Reals,
        1/2 (a-2 b+c)^2>=p*a^2+q*b^2+r*c^2
    ] && p>0 && q>0 && r>0,


  • $\begingroup$ Thank you! I will edit this further to try out my other hypotheses. $\endgroup$ – Cogicero Oct 30 '18 at 20:48
  • $\begingroup$ @Carl: What does False mean in this context? $\endgroup$ – Tugrul Temel Oct 31 '18 at 14:14
  • $\begingroup$ @Tugrul Based on my various "experiments" (a, b and c are vector norms), I think False means the expression cannot be reduced under the given constraints. $\endgroup$ – Cogicero Oct 31 '18 at 19:36

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