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.

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
$$
(1-a^2)\Delta=(1-a)^2(1+ay-bx)^2+(1-(a^2+b^2))(x-a)^2
$$
