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.

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.
