The inequality that I know under the name isodiametric inequality is
$$ \frac{\text{vol}(K)}{\text{diam}(K)^d} \le \frac{\text{vol}(B)}{\text{diam}(B)^d} $$
for any convex body $K$ in $\mathbb{R}^d$, where $B$ denotes the unit ball.
Proof 1: By Steiner symmetrization (which preserves volume, decreases diameter, and tends to the ball if desired). Proof 2: If $K$ has diameter at most 2 then $K-K\subseteq 2B$; by the Brunn-Minkowski inequality, $\text{vol}(K)\le\text{vol}(\frac12(K-K))$; thus $\text{diam}(K)\le\text{diam}(B)\implies \text{vol}(K)\le\text{vol}(B)$, which is equivalent to the desired inequality. Note that proof 2 doesn't really use the fact that $B$ is the Euclidean ball: it actually proves the more general analogous statement where we take $B$ to be any origin-symmetric convex body and measure diameters in the norm whose unit ball is $B$. (Proof 2 also yields that this isodiametric inequality is actually equivalent to the special case of Brunn-Minkowski that was used.) All of the above is in Gruber's recent book on convex geometry, for example.
Another proof of the generalization to arbitrary norms was given by M. S. Mel'nikov ("Dependence of volume and diameter of sets in an $n$-dimensional Banach space", Uspekhi Mat. Nauk 18(4) 165–170, 1963, http://mi.mathnet.ru/eng/umn6384): the key fact in that proof is that if the diameter of $K$ (in the sense of $B$) is at most 2 then the diameter of $K_t$ (in the sense of $B_t$) is also at most 2, where $K_t$ denotes the level set of height $t$ of the projection of $K$ (as a density) onto a fixed hyperplane; this allows a proof by induction on the dimension, and it anticipates the proof of the Prékopa-Leindler inequality, a generalization of Brunn-Minkowski. (For Prékopa-Leindler, see lecture 5 in Keith Ball's An Elementary Introduction to Modern Convex Geometry.)
Another inequality of the type you've asked about is Urysohn's inequality:
$$ \frac{\text{vol}(K)}{w(K)^d} \le \frac{\text{vol}(B)}{w(B)^d} $$
for any convex body $K$ in $\mathbb{R}^d$, where $B$ denotes the Euclidean unit ball and $w(\cdot)$ denotes mean width. (This time it really matters that it's the Euclidean ball.) Since $w(K)\le\text{diam}(K)$, this is a strengthening of the isodiametric inequality above.
Proof 1: Steiner symmetrization reduces mean width. Indeed, if $S_u$ denotes Steiner symmetrization wrt the hyperplane orthogonal to a unit vector $u$, and $R_u$ denotes reflection in that hyperplane, then $h_{S_u(K)}(\theta)=\frac12 h_K(\theta)+\frac12 h_K(R_u(\theta))$, where $h_K$ denotes the support functional of $K$; now integrate over $\theta\in S^{d-1}$ and use Jensen's inequality. (I got this from some unpublished notes of Giannopoulos.) Proof 2: See Pisier's book The Volume of Convex Bodies and Banach Space Geometry (Cambridge UP, 1989, p.6; Pisier writes that he learned this proof from Vitali Milman). In short, you generalize Minkowski addition of sets to Minkowski integration of set-valued functions, and you get an analogue of Brunn-Minkowski:
$$ \int_\Omega \text{vol}(A_t)^{1/n} \,d\mu(t) \le \text{vol}\left(\int_\Omega A_t \,d\mu(t)\right)^{1/n} $$
when $\mu$ is a probability measure and everything is suitably measurable. By symmetry, $\int_{O(d)} TK \,d\mu(T)$ is some multiple of the Euclidean ball (here $O(d)$ is the orthogonal group on $\mathbb{R}^d$, and $\mu$ is its Haar probability measure); a computation shows it's actually $\frac12 w(K)B$, and the Brunn-Minkowski analogue above finishes the proof.
As requested in comments, here's a generalization to other intrinsic volumes:
$$ 1\le i\le j\le d\implies
\frac{V_i(B)^{1/i}}{V_j(B)^{1/j}}
\le \frac{V_i(K)^{1/i}}{V_j(K)^{1/j}}$$
(The case $i=1$, $j=d$ is Urysohn's inequality.) Proof: A special case of the Alexandrov-Fenchel inequality is
$$ W_i(K)^2 \ge W_{i-1}(K) W_{i+1}(K) \tag{$\ast$} $$
where $W_i(\cdot)$ denotes quermassintegrals:
$$ W_i(K) = V(\underbrace{K,\dotsc,K}_{d-i},\underbrace{B,\dotsc,B}_i)
= \frac{\kappa_i}{\binom di} V_{d-i}(K) $$
where $\kappa_i$ is the volume of the $i$-dimensional unit Euclidean ball. It follows that
$$ i\mapsto\left(\frac{W_d(K)}{W_{d-i}(K)}\right)^{1/i} \tag{$\dagger$} $$
is an increasing function for $1\le i\le d$. (You can just prove the $i$-vs-$(i+1)$ case by induction on $i$, but what's really going on here is that $i\mapsto\log W_i(K)$ is "concave" — scare quotes because its domain is discrete. The inequality ($\ast$) is the local version of this, analogous to saying that the second derivative is nonpositive; that ($\dagger$) is increasing means that the slopes over $[d-i,d]$ are increasing with $i$.) But $W_d(K) = \text{vol}(B) = W_{d-i}(B)$, so a bit of rearrangement yields the desired inequality.
(Unfortunately I'm not familiar with the literature around Alexandrov-Fenchel, so I can't give good references here.)
You might also want to consider things like the reverse isoperimetric inequality, which asserts that (1) every centrally symmetric convex body $K$ has an affine image $K'$ such that
$$ \frac{V_d(K')^{1/d}}{V_{d-1}(K')^{1/(d-1)}} \ge \frac{V_d(B_\infty^d)^{1/d}}{V_{d-1}(B_\infty^d)^{1/(d-1)}} $$
where $B_\infty^d$ is the cube $[-1,1]^d$ (i.e., the unit ball of the $\ell_\infty^d$ norm), and that (2) every convex body $K$ has an affine image $K'$ such that
$$ \frac{V_d(K')^{1/d}}{V_{d-1}(K')^{1/(d-1)}} \ge \frac{V_d(\Delta)^{1/d}}{V_{d-1}(\Delta)^{1/(d-1)}} $$
These inequalities are due to Keith Ball (see lecture 6 of his book mentioned above for the proof and references), relying on John's theorem and a normalized version of the Brascamp-Lieb inequality. For a proof of Brascamp-Lieb in the needed form (and much more besides, including equality cases in the above reverse isoperimetric inequalities), see F. Barthe, "On a reverse form of the Brascamp-Lieb inequality", arxiv:math/9705210. (A simplified version for the needed one-dimensional special case appears in K. Ball, "Convex geometry and functional analysis", in volume 1 of Handbook of the Geometry of Banach Spaces, Johnson and Lindenstrauss (eds.), North-Holland, 2001.)
We follow the analogous development in this answer. Let $S(D)$ be given by
$$\begin{align}
S(D)&=\sum_{n=-\infty}^\infty \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{|2n+1|}\right)}{n+3/4}\\\\&=\sum_{n=0}^\infty \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}+\sum_{n=-\infty}^{-1} \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{|2n+1|}\right)}{n+3/4}\\\\
&=\sum_{n=0}^\infty \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}+\sum_{n=0}^\infty \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+1/4}\tag1
\end{align}$$
We analyze the first series on the right-hand side of $(1)$. We begin by writing
$$\begin{align}
\sum_{n=0}^\infty \frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}&=\sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}\\\\
&+\sum_{2n+1> D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}
\end{align}\tag2$$
For the first series on the right-hand side of $(2)$ we have
$$\begin{align}
\sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}&=\log(D)\sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)}{n+3/4}\\\\
&-\sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(2n+1\right)}{n+3/4}\tag3
\end{align}$$
For the first series on the right-hand side of $(3)$ we find using the Euler-McLaurin Summation Formula that
$$\begin{align}
\sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)}{n+3/4}&=\frac\pi2 \sum_{2n+1\le D}\frac1{n+3/4}-\sum_{2n+1\le D}\frac{\arctan\left(\frac {2n+1}{D}\right)}{n+3/4}\\\\
&=\frac\pi2\left(\log(D)+O(1)\right)-O(1)\\\\
&=\frac\pi2 \log(D)+O(1)\tag4
\end{align}$$
For the second series on the right-hand side of $(3)$ we find using the Euler-McLaurin Summation Formula that
$$\begin{align}
\sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log(2n+1)}{n+3/4}&=\frac\pi2 \sum_{2n+1\le D}\frac{\log(2n+1)}{n+3/4}\\\\
&-\sum_{2n+1\le D}\frac{\arctan\left(\frac{2n+1}{D}\right)\log(2n+1)}{n+3/4}\\\\
&=\frac\pi4 \log^2(D)+O\left(\frac{\log(D)}{D}\right)-O(1)\tag5
\end{align}$$
Using $(4)$ and $(5)$ in $(3)$ reveals
$$\sum_{2n+1\le D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}=\frac\pi4 \log^2(D)+O(\log(D))$$
Next, we anlayze the second series on the right-hand side of $(2)$. It is evident that
$$\left|\sum_{2n+1> D}\frac{\arctan\left(\frac {D}{2n+1}\right)\log\left(\frac {D}{2n+1}\right)}{n+3/4}\right|\le D^2\sum_{2n+1>D}\frac{1}{(2n+1)^2(n+3/4)}=O(1)$$
Putting it all together, we find that for $D\to\infty$
$$S(D)=\frac{\pi}{2}\log^2(D)+O(\log(D))$$
Best Answer
Some thoughts.
According to my answer, given $s, A > 0$ with $s^4 \ge 27A^2$, the minimum of $a + b - c$ is given by the smallest positive real root $v_1 = v_1(s, A)$ of the cubic equation $$sv(2s - v)^2 - 32A^2 = 0. \tag{1}$$
Let $w = \frac{v_1}{8s}$ and $m = \frac{A^2}{s^4} \in (0, 1/27]$. (1) is written as $$w(1 - 4w)^2 = m. \tag{2}$$
Using trigonometric solutions of the cubic (2), we have $$w = \frac16 - \frac16 \sin \left(\frac{\pi}{6} + \frac13 \arccos(54 m - 1)\right). \tag{3}$$
Now calculate the Maclaurin series of $w = w(m)$. We have $$m+8\,{m}^{2}+112\,{m}^{3}+1920\,{m}^{4}+36608\,{m}^{5}+745472\,{m}^{6} +15876096\,{m}^{7} + \cdots \tag{4}$$
We see that $C_1 = 1, C_2 = 8, C_3 = 112, C_4 = 1920$ etc.