2
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

The following shows that Reduce thinks there is no answer:

Reduce[
    ForAll[
        {a,b,c},
        {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,
    {p,q,r}
]

False

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.