I want to show that the polynomial,
$$(2c+1)(b-2c)(b-2c+d-1)(a/2+d-2c)(2+a/2-d+2c)-(b-2c+2d-1)(2c+1-d)^2$$
is greater than zero, for $a,b,c,d \in \mathbb Z$ subject to the constraints:
- $a \mod 4 = 0$
- $b \geq a/2$
- $c \leq a/4-1$
- $d \leq 2c$
and note $a,b,c,d$ are bounded from below by zero. Since this isn't a real polynomial, I can't just prove the non-existence of real roots to establish that the polynomial $> 0$.
I would appreciate any help, but please do not post a full solution. Note also this arose from original work, and the form the polynomial is written is thus not any hint to a solution.
If I am not mistaken, one observation I made that may be useful is given the constraints none of the terms in brackets are ever negative.
$$(2c+1)(b-2c)(b-2c+d-1)(a/2+d-2c)(2+a/2-d+2c)-(b-2c+2d-1)(2c+1-d)^2$$
We want $$(2c+1)(b-2c)(b-2c+d-1)(\frac{a}{2}+d-2c)(2+\frac{a}{2}-d+2c) \geq(b-2c+2d-1)(2c+1-d)^2$$
as $ c \leq \frac{a}{4} -1,$ we get $ 2c \leq \frac{a}{2} -2,$ with $ -2c \geq 2- \frac{a}{2} .$ Since $ b \geq \frac{a}{2} .$ we reach $$ b-2c \geq 2. $$ It follows that $$(b-2c)(b-2c+d-1) \geq 2(b-2c) + 2d - 2, $$ $$(b-2c)(b-2c+d-1) \geq (b-2c) + 2d + (b-2c) -2 , $$ $$(b-2c)(b-2c+d-1) \geq (b-2c) + 2d \geq (b-2c+2d-1). $$ $$ \color{blue}{(b-2c)(b-2c+d-1) } \geq (b-2c+2d-1). $$
Next we see $$ \color{blue}{ (2c+1)} \geq (2c+1 - d), $$ $$ \color{blue}{ (2+\frac{a}{2}-d+2c) } \geq (2c+1 - d). $$
Mix in our three inequalities, we find $$(2c+1)(b-2c)(b-2c+d-1)(\frac{a}{2}+d-2c)(2+\frac{a}{2}-d+2c) \geq(b-2c+2d-1)(2c+1-d)^2$$