Determining whether an inequality is satisfied Let $a \geq 1$ and $b \geq 1$. Let's pick two vectors $\mathbf{x}$ and $\mathbf{y}$ such that:
$$
\mathbf{x} \cdot \mathbf{x} = a^2 + 1 \\
\mathbf{y} \cdot \mathbf{y} = b^2 - 1 \\
$$
Is there a way to determine a bound on $\mathbf{x} \cdot \mathbf{y}$?
Specifically, I'd like to show that the combination $\mathbf{x} \cdot \mathbf{y} + a b > 0$, but I can't quite figure it out.
I know that the inner product is bound by the values $-|\mathbf{x}||\mathbf{y}| \leq \mathbf{x} \cdot \mathbf{y} \leq |\mathbf{x}||\mathbf{y}|$, but I can't quite figure out how to use this.
 A: So we have $||\vec{x}||_2^2 = a^2 + 1,\ ||\vec{y}||_2^2 = b^2 - 1$ and it's also clear that $ab > 0$. 
So if $\vec{x} \cdot \vec{y} \geq 0$ then the inequality is trivially true since every term are positive.
So now let's suppose that $\vec{x} \cdot \vec{y} \leq 0$. By C-S we know :
$$
\begin{align}
 -\vec{x} \cdot \vec{y} &\leq ||\vec{x}||_2||\vec{y}||_2 \\
0 &\leq ||\vec{x}||_2||\vec{y}||_2 + \vec{x} \cdot \vec{y}= \vec{x} \cdot \vec{y} + \sqrt{(a^2+1)(b^2-1)}
\end{align}
$$
So now we would need to show that $\sqrt{(a^2+1)(b^2-1)} \leq ab$. However this is wrong for arbitrary $a,b \geq 1$. So let's try to derive a counter-example. .
Set $$\vec{x} := (5,5),\ \vec{y} := (-6,-6) \implies ||\vec{x}||_2^2 = 50,\ ||\vec{y}||_2^2 = 72 \implies$$
$$ a = \sqrt{50-1} = 7,\ b = \sqrt{72+1} \approx 8.54$$
Then 
$$
\vec{x} \cdot \vec{y} + ab \approx -60 + 59.8 < 0
$$
A: Geometric viewpoint: If $\alpha$ is the angle between the vectors
$$\mathbf{x}\cdot \mathbf{y} = \|\mathbf{x}\|\|\mathbf{y}\|\cos\alpha$$
and $|\mathbf{x}\cdot\mathbf{y}|\le\|\mathbf{x}\|\|\mathbf{y}\| = \sqrt{(a^2+1)(b^2-1)}$, But as @Zubzub has already said, your required inequality is trivial for $\mathbf{x}\cdot \mathbf{y}\ge 0$  and false in the other case.
