# 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.

• Yah, It is kind of CS inequality! Thanks. (BTW, use CS instead of the spelling, it confuses me often) – Swapnil Das Dec 21 '15 at 10:17
• @G-man * Schwarz – Prince M Dec 21 '15 at 10:20
• Maybe, I'm new to CS, so it isn't as obvious to me as it maybe to you, Genius :) – Swapnil Das Dec 21 '15 at 10:20
• But the $\ge$ is in the opposite direction as in CS inequality... – Jimmy R. Dec 21 '15 at 10:20
• Maybe we could work on it and shift it to the opposite :P – Swapnil Das Dec 21 '15 at 10:21

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 $\langle a, a \rangle \cdot \langle x, x \rangle- \langle a, x \rangle^2$

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

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

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.