Does every triangle satisfy $(a+b-c) \ge 8s\sum_{k=1}^{\infty}C_k \left(\frac{A}{s^2}\right)^{2k}$?

Question

For a fixed semi-perimeter $s$ and a fixed area $A$, we can find infinitely many triangles of sides $a,b$ and $c$ whose semi-perimeter is $s$ and area is $A$. I wanted to improve the triangle inequality by finding a expression for the minimum value of $a+b-c$ in terms of $s$ and $A$. Partial progress was made in this question and also this question.

I obtained the following inequality experimentally using Monte Carlo simulation.

$$ a+b-c \ge \frac{8A^2}{s^3} + \frac{64A^4}{s^7} + \frac{896A^6}{s^{11}} $$

The first term of $(1)$ is proved in first of the above links. $(1)$ suggests the folowing conjecture:

Conjecture: There exists constants $C_k$ such that
$$ \min(a+b-c) = 8s\sum_{k=1}^{\infty}C_k \left(\frac{A}{s^2}\right)^{2k}\tag 1 $$

Can this conjecture be proved and can we express $C_k$ in terms of $k$? We have $C_1 = 1, C_2 = 8$ and $C_3 = 112$. Interestingly, there are several sequences in OEIS which begin with $1,8,112$.

SageMath Code for estimating $C_4$

import random as rn

R  = 1
x1 = 0
y1 = R
min1 = 10**999
stop = False
i = 1

while stop == False:
    x2 = R*rn.random()
    y2 = (R^2 - x2^2)^0.5
    x3 = R*rn.random()
    y3 = (R^2 - x3^2)^0.5
    a = (((x2)**2 + (y1 - y2)**2)**0.5)
    b = (((x2 - x3)**2 + (y2 - y3)**2)**0.5)
    c = (((x3 - x1)**2 + (y3 - y1)**2)**0.5)
    s    = ((a+b+c)/2)
    area = ((s*(s-a)*(s-b)*(s-c))**0.5)
        
    try:
        test1 = ((a+b-c) - (8*area^2/s^3 + 64*area^4/s^7 + 896*area^6/s^11)) / (area^8/s^15)
    except:
        continue

    if test1 < min1:
        min1 = test1
        print(a,b,c,min1/8)
    
        if min1 <= 0:
            print('exception', min1.n(), p)
            stop = True
            break
            
    if i%10^7 == 0:
        print(i)
    i = i + 1

Answer 1

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.