Verifying a Stubborn Inequality 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.  
 A: 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}
A: 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.
