Proving $(ax+by-1)^2 \ge (x^2+y^2-1)(a^2+b^2-1)$

Question

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) — Das, Dec 21 '15 at 10:17
Maybe, I'm new to CS, so it isn't as obvious to me as it maybe to you, Genius :) — 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

score 2 · Answer 1 · answered Dec 21 '15 at 10:28

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

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

answered Dec 21 '15 at 10:28

Ewan Delanoy

61,600

score 0 · Answer 2 · answered Dec 21 '15 at 11:05

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.

score 0 · Accepted Answer · answered Dec 21 '15 at 11:43

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

Proving $(ax+by-1)^2 \ge (x^2+y^2-1)(a^2+b^2-1)$

3 Answers3