Is this a stronger form of the triangle inequality?

Question

Let $a,b,c$ be the length of the sides of a triangle, $s$ be its semi-perimeter and $A$ be its area. I wanted to see if the well know Euclidean triangle inequality $a+b > c$ can be improved. I inscribed billions of triangles of all shapes in a unit circle centered on the origin (code given below) and observed

$$ a + b - c \ge \frac{8A^2}{s^3}\bigg(1 + \frac{4A^2}{c^4}\bigg) \tag 1 $$

Since both terms on the RHS are positive, the standard triangle inequality follows. The first term is a constant for a fixed triangle while the second term is dependent on $c$. Intuitively, this makes sense as it shows that $a+b-c$ not only depends on the triangle but also on which side is selected from the other two. I could not find this inequality my searches so I am not sure if this is known or not.

Question: I am looking for a proof or any reference in existing literature.

import random as rn

n = 1
min = 10**9

while True:
    x1 = ((-1)**rn.randint(0,1))*(rn.random())
    y1 = ((-1)**rn.randint(0,1))*(1 - x1**2)**0.5
    x2 = ((-1)**rn.randint(0,1))*(rn.random())
    y2 = ((-1)**rn.randint(0,1))*(1 - x2**2)**0.5
    x3 = ((-1)**rn.randint(0,1))*(rn.random())
    y3 = ((-1)**rn.randint(0,1))*(1 - x3**2)**0.5
    
    a = round(((x1 - x2)**2 + (y1 - y2)**2)**0.5,14)
    b = round(((x2 - x3)**2 + (y2 - y3)**2)**0.5,14)
    c = round(((x3 - x1)**2 + (y3 - y1)**2)**0.5,14)
    
    if (a+b < c) or (b+c < a) or (c+a < b):
        print(x1,y1,x2,y2,x3,y3,a,b,c)
        break
    
    s = round((a+b+c)/2,14)
    area = round((s*(s-a)*(s-b)*(s-c))**0.5,14)
    
    # Test inequality
    test = round((((a+b-c)*(s**3)/(area**2) - 8)*(c**4)/(area**2)),14)
    
    if test < min:
        min = test
        print('Inequality',a, b, c, s, area, min)
    
    if n%1000000 == 0:
        print(n)
    
    n = n + 1

Answer 1

Substitute $a+b-c=2(s-c)$ and $A^2=s(s-a)(s-b)(s-c)$:

$$2(s-c)\ge \frac{8s(s-a)(s-b)(s-c)}{s^3}\left(1+\frac{4s(s-a)(s-b)(s-c)}{c^4}\right).$$

Cancel out $2(s-c)$:

$$1\ge \frac{4(s-a)(s-b)}{s^2}\left(1+\frac{4s(s-a)(s-b)(s-c)}{c^4}\right) = (\ast).$$

Note that by AM-GM $4(s-a)(s-b) \le ((s-a)+(s-b))^2 = c^2$, hence

$$(\ast) \le \frac{c^2}{s^2}\left(1+\frac{s(s-c)}{c^2}\right)=1-\frac{c(s-c)}{s^2} \le 1.$$

Done!