Numerical methods to minimize a maximum Let $r(t)$ be the function:
$r(t) = \sqrt{x(t)^2 + y(t)^2}$, where
$x(t) = 3b (1 − t)^2 t + 3c (1 − t) t^2 + a t^3$, and
$y(t) = a (1 − t)^3 + 3c (1 − t)^2 t + 3b (1 − t) t^2$
Now define $f$:
$f(a, b, c) = \max( |r(t) - 1|, 0 \leq t \leq 1)$
Based on the plots I've made, $f$ should have a local minimum near:
$(a, b, c) = (1.0, 0.5, 1.0)$
But I'm struggling to find a way to numerically calculate it.
Ideally, I'd like to solve this using Wolfram Alpha:
https://www.wolframalpha.com/
Or Maxima:
https://maxima.sourceforge.io/download.html
But I'd be grateful if you know of any way to do it!
I've spent almost a full day on this. Everything I've tried is either interpreted incorrectly, or eventually fails due to a timeout.
 A: 7/22/2022 Update:
I rewrote this algorithm to achieve more significant figures without further alteration.  However, it likely takes at least several minutes on a typical computer, and I wouldn't claim that it achieves all 9-12 significant figures sought (4-5 is a much more reasonable claim).  I wasn't aware of the number of significant figures sought when I first wrote an answer.  The algorithm gives values of a = 1.000052922, b = 0.553420099, and c = 0.998742294.
I appreciate the comments regarding my approach to this problem.
I did attempt to write a "Hooke & Jeeves-like" algorithm in R based on feedback from @PierreCarre and a general description of the algorithm found online.  I found that even modest changes in the value of a,b, and c can alter the shape of $|r(t) - 1|$ across the interval for $t$, implying that the maximum and corresponding minimums also change.  When I tried varying the coefficients one at a time, the minimum of $f$ did not always decrease.  Another challenge is that typical optimization routines for Nelder-Meade and similar algorithms in R most commonly minimize a function based on the independent variable rather than the coefficients, which makes their use less obvious for problems like this.
Regarding the comment of @Spencer, the algorithm can only find a,b, and c that yield the lowest value of $f$ in the search space.  Outside of that, I wouldn't claim to provide any guarantees of achieving a local minimum because the shape of $f$ can vary so much based on the values of a,b, and c.
With the previously mentioned caveats regarding the runtime and the limited number of significant figures, I will leave this algorithm online in case someone finds it useful.
#Written in R 4.2.1
require(data.table)
##define functions
x = function(t,a = aa,b = bb,c = cc) 3*b*(1-t)^2*t + 3*c*(1-t)*t^2 + a*t^3
y = function(t,a = aa,b = bb,c = cc) a*(1-t)^3 + 3*c*(1-t)^2*t + 3*b*(1-t)*t^2
rt = function(t,a,b,c) sqrt(x(t,a,b,c)^2 + y(t,a,b,c)^2)
f = function(t,a,b,c) max(abs(rt(t,a,b,c) - 1))
 
avec = NULL;bvec = NULL; cvec = NULL
granvec = c(1e-2,1e-3,1e-4,1e-5,1e-6,1e-7,1e-8,1e-9)
fansvec = NULL
a = 1;b = .55;c = 1
sn = 20  ##number to define a,b,c sequences
 
for(j in 1:length(granvec)){
gran = granvec[j]   #granularity of the search space
aa = seq(a - sn*gran,a + sn*gran,gran)
bb = seq(b - sn*gran,b + sn*gran,gran)
cc = seq(c - sn*gran,c + sn*gran,gran)
tt = seq(0,1,1e-4)
#data frame of all combinations in search space to specified granularity
sspace2 = CJ(aa,bb,cc)

ansframe2 = NULL
for(i in 1:dim(sspace2)[1]){
ansframe2[i] = f(tt, a = sspace2$aa[i], b = sspace2$bb[i], c = sspace2$cc[i])
}

sspace2$fans = ansframe2
sspace2[sspace2$fans == min(sspace2$fans),]

fansvec[j] = sspace2[sspace2$fans == min(sspace2$fans),]$fans[1]
 
if(j > 1){
if(fansvec[j] > fansvec[j-1]){
print("Algorithm is not converging to the true local minimum")
}}
a = sspace2[sspace2$fans == min(sspace2$fans),]$aa[1]
b = sspace2[sspace2$fans == min(sspace2$fans),]$bb[1]
c = sspace2[sspace2$fans == min(sspace2$fans),]$cc[1]
avec[j] = a;bvec[j] = b;cvec[j] = c
}

ansvec = data.frame(avec,bvec,cvec)
options(digits = 15)
f(tt,a = a,b = b,c = c)
[1] 5.55341550056987e-05
a
[1] 1.000052922
b
[1] 0.553420099
c
[1] 0.998742294

A: Some thoughts:
Since $r(t) = r(1 - t)$ for all $t\in [0, 1]$, we consider the following equivalent problem:
$$\min_{a, b, c} ~ \max_{t\in [0, 1/2]} ~ [r(t) - 1]^2. $$
With $t \in [0, 1/2]$, let $s = (1/2 - t)^2 \in [0, 1/4]$ (correspondingly $t = 1/2 - \sqrt s$).
The optimization problem becomes
$$\min_{a, b, c} ~ \max_{s \in [0, 1/4]} ~ (\sqrt{As^3 + Bs^2 + Cs + D} - 1)^2$$
where
\begin{align*}
 A &= 2(a + 3b - 3c)^2, \\
 B &= \frac{15a^2 - 6ab - 30ac - 9b^2 + 54bc - 9c^2}{2}, \\
 C &= \frac{15a^2 - 6ab + 30ac - 9b^2 - 54bc - 9c^2}{8}, \\
 D &= \frac{(a + 3b + 3c)^2}{32}.
\end{align*}
Let $$g(s) := (\sqrt{As^3 + Bs^2 + Cs + D} - 1)^2.$$
We have $g(0) = [(a + 3b + 3c)\sqrt 2/8 - 1]^2$ and $g(1/4) = (|a| - 1)^2$.
We have
$$g'(s) = 2(\sqrt{As^3 + Bs^2 + Cs + D} - 1)\cdot \frac{3As^2 + 2Bs + C}{2\sqrt{As^3 + Bs^2 + Cs + D}}.$$
(Note: At maximum, $\sqrt{As^3 + Bs^2 + Cs + D} - 1 \ne 0$. So, we only need to consider $3As^2 + 2Bs + C = 0$.)
So we can solve $\max_{s \in [0, 1/4]} ~ (\sqrt{As^3 + Bs^2 + Cs + D} - 1)^2$ in closed form.
Moreover, based on the existing answer, I believe in the neighborhood of $a = 1.000052922, b = 0.553420099, c = 0.998742294$, the global maximizer $s^\ast$ is one of the roots of $3As^2 + 2Bs + C = 0$.
