How to prove a multi-variable polynomial inequality (always positive or negative)?

I encountered the following polynomial in a proof of a geometry problem, which should be always negative: $$b^5c^4+2b^5c^3d+b^5c^2d^2+b^4c^5+6b^4c^4d+5b^4c^3d^2-4b^4c^3-4b^4c^2d+2b^3c^5d+5b^3c^4d^2-4b^3c^4-14b^3c^3d-5b^3c^2d^2+5b^3c^2+4b^3cd-b^3+b^2c^5d^2-4b^2c^4d-5b^2c^3d^2+5b^2c^3+14b^2c^2d+4b^2cd^2-5b^2c-2b^2d+4bc^3d+4bc^2d^2-5bc^2-6bcd-bd^2-c^3-2c^2d-cd^2$$ Under the condition that $|d|<\min(b,c)$, $b<1$, $c<1$ and $b\neq c$. Because of the symmetry of b and c, we can think $|d|<c<b<1$.

It is obvious for $d = 0$ because it can be simplified as $(b+c)(b^4c^4-4b^3c^3+5b^2c^2-b^2-4bc-c^2)$, and the second factor is: $$b^4c^4-4b^3c^3+5b^2c^2-b^2-4bc-c^2=\\b^3c^3(bc-1)-3b^3c^3+5b^2c^2-(b+c)^2-2bc<\quad\left[(b+c)^2>4bc\right]\\b^3c^3(bc-1)-3b^3c^3+5b^2c^2-6bc=\\b^3c^3(bc-1)-3b^3c^3+5bc(bc-1)-bc<0\quad\left[bc-1<0,\,bc>0\right]\\$$

But I don't know how to deal with the additional variable $d$.

The inequity can be numerically checked:

def main():
    for i in range(0, 11):
        for j in range(0, i + 1):
            for k in range(-j, j + 1):
                b, c, d = i*.1, j*.1, k*.1
                p = b**5*c**4 + 2*b**5*c**3*d + b**5*c**2*d**2 + b**4*c**5 + 6*b**4*c**4*d + 5*b**4*c**3*d**2 - 4*b**4*c**3 - 4*b**4*c**2*d + 2*b**3*c**5*d + 5*b**3*c**4*d**2 - 4*b**3*c**4 - 14*b**3*c**3*d - 5*b**3*c**2*d**2 + 5*b**3*c**2 + 4*b**3*c*d - b**3 + b**2*c**5*d**2 - 4*b**2*c**4*d - 5*b**2*c**3*d**2 + 5*b**2*c**3 + 14*b**2*c**2*d + 4*b**2*c*d**2 - 5*b**2*c - 2*b**2*d + 4*b*c**3*d + 4*b*c**2*d**2 - 5*b*c**2 - 6*b*c*d - b*d**2 - c**3 - 2*c**2*d - c*d**2
                print(f'{p:.4f} = P({b:.1f},{c:.1f},{d:.1f})')
                if p > 0:
                    return 

if __name__ == '__main__':
    main()

Backgroud

I'm trying the coordinate proof of this problem, and here is my trying:

$$0<\angle ABC=2\beta<\pi/2\\0<\angle ACB=2\gamma<\pi/2\\0<\angle DBC=\beta+\delta\\0<\angle DBA=\beta-\delta\\0<\angle ECB=\gamma+\delta\\0<\angle EBA=\gamma-\delta \\|δ|<min(\beta,\gamma)$$

When given these angles, we have: $$\angle DBC+\angle ECA=\angle DBA+\angle ECB=\beta+\gamma$$ which follows $\angle 1+\angle 2=\angle 3+\angle 4$. So the task is to prove $\beta=\gamma$ by using $BD=CE$. (Note that if $\delta=0$, it's Steiner-Lehmus theorem.)

When coordinate the triangle, I put $A$ onto y-axis and $BC$ onto x-axis, then I have: $$A=(0,a)\\B=(-\frac{a}{\tan2\beta},0)\\C=(\frac{a}{\tan2\gamma},0)$$

To make all coordinates rational, we make:

$$0<b=\tan\beta<1\\0<c=\tan\gamma<1\\d=\tan\delta,|d|<\min(b,c)$$

Then we get all coordinates:

$$A=\left(0,a\right)\\B=\left(-\frac{a(1-b^2)}{2b},0\right)\\C=\left(\frac{a(1-c^2)}{2c},0\right)\\D=\left(\frac{a(b-d)(b^2+1)(c-1)(c+1)}{2b(bc^2+2bcd-b+c^2d-2c-d)},\frac{a(b+c)(b+d)(bc-1)}{b(bc^2+2bcd-b+c^2d-2c-d)}\right)\\ E=\left(-\frac{a(b-1)(b+1)(c-d)(c^2+1)}{2c(b^2c+b^2d+2bcd-2b-c-d)},\frac{a(b+c)(c+d)(bc-1)}{c(b^2c+b^2d+2bcd-2b-c-d)}\right)$$

Then the condition $BD=CE$ can be written as:

$$BD^2-CE^2=\frac{a^2(b-c)(b+c)^2(d^2+1)(bc-1)^2(b^5c^4+2b^5c^3d+b^5c^2d^2+b^4c^5+6b^4c^4d+5b^4c^3d^2-4b^4c^3-4b^4c^2d+2b^3c^5d+5b^3c^4d^2-4b^3c^4-14b^3c^3d-5b^3c^2d^2+5b^3c^2+4b^3cd-b^3+b^2c^5d^2-4b^2c^4d-5b^2c^3d^2+5b^2c^3+14b^2c^2d+4b^2cd^2-5b^2c-2b^2d+4bc^3d+4bc^2d^2-5bc^2-6bcd-bd^2-c^3-2c^2d-cd^2)}{b^2c^2(bc^2+2bcd-b+c^2d-2c-d)^2(b^2c+b^2d+2bcd-2b-c-d)^2}$$

I should prove this complicated expression be zero only if $b=c$, so the last long factor on numerator should be proved non-zero.

Proof:

Let $f(b, c, d)$ denote the expression.

We need to prove that $f(b, c, d) < 0$ for all real numbers $b, c, d$ satisfying $|d| < c < b < 1$.

The condition $|d| < c < b < 1$ can be transformed to \begin{align*} &b = 1 - \frac{1}{1 + s + t}, \\ &c = 1 - \frac{1}{1 + s}, \\ &d = (-c)\cdot \frac{1}{1 + r} + c \cdot \frac{r}{1 + r}, \\ &s, t, r > 0. \end{align*} (Correspondingly, $r = \frac{c + d}{c - d}, \, s = \frac{c}{1 - c},\, t = \frac{b - c}{(1 - b)(1 - c)}$.)

Then we have $$f(b, c, d) = - \frac{g(s, t, r)}{(1 + s + t)^5(1 + s)^7(1 + r)^2}$$ where $g(s, t, r)$ is a nonzero polynomial with non-negative coefficients (a long expression).

We are done.