Proving $(ax+by-1)^2 \ge (x^2+y^2-1)(a^2+b^2-1)$ 
Let $x$, $y$, $a$, $b$ be real numbers such that $a^2+b^2 \leq 1$ and $x^2+y^2 \leq 1$. Show that $$(ax+by-1)^2 \ge (x^2+y^2-1)(a^2+b^2-1)$$

I am unable to find a solution to this problem. My initial thoughts were to have a trigonometric substitution of variables, but that didn't lead me further. Please help.
Thank you.
 A: This is a quadratic inequality in $x,y$, so all you have to do
is complete the squares.
Formally, if $\Delta=(ax+by-1)^2 -(x^2+y^2-1)(a^2+b^2-1)$ then you
have
$$
(1-a^2)\Delta=(1-a)^2(1+ay-bx)^2+(1-(a^2+b^2))(x-a)^2
$$
A: Use C-S we have
$$(ax+by)^2\le (a^2+b^2)(x^2+y^2)\le 1
\Longrightarrow ax+by\le 1$$
we can rewrite the inequality 
$$(1-ax-by)^2\ge (1-a^2-b^2)(1-x^2-y^2)$$
$$\Longleftrightarrow 1-(ax+by)\ge \sqrt{(1-a^2-b^2)(1-x^2-y^2)}$$
Let $A=a^2+b^2,B=x^2+y^2$,since
$$1-(ax+by)\ge 1-\sqrt{AB}$$it is enought to show that
$$1-\sqrt{AB}\ge \sqrt{(1-A)(1-B)}$$
Use C-S
we have
$$1=[(1-A)+A][(1-B)+B]\ge(\sqrt{(1-A)(1-B)}+\sqrt{AB})^2$$
By done.
A: HINT:
This should be the C-S inequality for Lorentzian metrics. Assume that
$a_0^2 - \sum_{k=1}^n a_k^2 > 0$, $x_0^2 - \sum_{k=1}^n x_k^2 > 0$   and let's show that 
$$(a_0 x_0 - \sum_{k=1}^n a_k x_k )^2 \ge (a_0^2 - \sum_{k=1}^n a_k^2)(x_0^2 - \sum_{k=1}^n x_k^2)$$
or
$$\langle a, x \rangle^2 \ge \langle a, a \rangle \cdot \langle x, x \rangle$$
for the Lorentzian bilinear form $\langle a, x\rangle = a_0 x_0 - \sum_{k=1}^n a_k x_k$, of signature $(1,n)$.  
Assume that the vectors $(a_0, \ldots, a_n)$ and $(x_0, \ldots x_n)$ are not proportional. Consider the quadratic form on $\mathbb{R}^2$, 
$$q((u,v)) = \langle u a + v x, u a + v x\rangle$$
We have $q(1,0) = \langle a,  a \rangle>0$ ( and same for $x$). Now, the Lorentzian quadratic form is negative definite on the subspace $0 \times \mathbb{R}^n$ and that subspace will contain a non-zero $u a+ v x$. For that $(u,v)$ we have $q(u,v) < 0$. We conclude that the signature of the form $q$ is $(1,1)$ and so its determinant is $< 0$. But that determinant is exactly $\langle a, a \rangle \cdot \langle x, x \rangle- \langle a, x \rangle^2$
