Verifying a Stubborn Inequality

Question

I have an inequality which arose in an unrelated problem, and it's been proving to get the better of me. I essentially know it has to be true, but cannot verify it. For integers $n_{1}, n_{2}, \ell_{1}, \ell_{2}$, I want to show that

$$8 n_{1} + 8n_{2} -\ell_{1}^{2} - \ell_{2}^{2} - 2\ell_{1} \ell_{2} > 0$$

Given only the constraints $4n_{1} - \ell_{1}^{2} \geq -1$, $4n_{2} - \ell_{2}^{2} >0$, $n_{1} \geq 0$, and $n_{2} > 0$. I am also happy to assume $\ell_{1} + \ell_{2} >0$, and that $\ell_{1} \geq \ell_{2} >0$. In my particular problem, this case will suffice. Does anyone have any tips on proving this?

I realize (given the form of the constraints) it's tempting to rewrite the lefthand side as

$$(4n_{1} -\ell_{1}^{2})+(4n_{2} - \ell_{2}^{2}) + 4n_{1} + 4n_{2} - 2\ell_{1} \ell_{2},$$

but the term $-2 \ell_{1} \ell_{2}$ is troubling, as it is negative by my assumptions.

Answer 1

With less complicated symbols, you want to prove that $8a+8b-x^2-y^2-2xy>0$ provided $4a-x^2\ge-1$ and $4b-y^2>0$.

First notice that $4b-y^2>1$, because $4b-y^2=1$ implies $y^2\equiv 3\pmod{4}$, which is impossible. Now rewrite the expression as \begin{align} 2(4a-x^2)+2(4b-y^2)+x^2+y^2-2xy &=2(4a-x^2)+2(4b-y^2)+(x-y)^2\\ &>-2+2+0=0 \end{align}

Answer 2

You can rewrite your inequality as

$$2\underbrace{(4n_1 - \ell_1^2)}_{\ge -1} + 2\underbrace{(4n_2 - \ell_2^2)}_{\ge 1} + \underbrace{(\ell_1 - \ell_2)^2}_{\ge 0} \;\stackrel{?}{>}\; 0$$

It is easy to see LHS $\ge 2(-1)+2(1) + 0 = 0$. What we need to do is rule out the possibility LHS $= 0$. In order for LHS $= 0$, we need

$$4n_1 - \ell_1^2 = -1,\quad 4n_2 - \ell_2^2 = 1,\quad\text{ and }\quad \ell_1 - \ell_2 = 0$$ If $\ell_1 = \ell_2$, then subtracting the first equation from second, we get $$4(n_2 - n_1) = (4n_2 - \ell_2^2) - (4n_1 - \ell_1^2) = 1 - (-1) = 2$$ which is clearly impossible. This implies LHS $\ne 0$ and the desired inequality LHS $> 0$ follows.