Prove that $(x_1y_1 + x_2y_2 - 1)^2 \ge (x_1^2 + x_2^2 - 1)(y_1^2 + y_2^2 - 1)$

Question

$x_1, y_1, x_2, y_2 \in \mathbb R$

$x_1^2 + x_2^2 \le 1$

Prove that $(x_1y_1 + x_2y_2 - 1)^2 \ge (x_1^2 + x_2^2 - 1)(y_1^2 + y_2^2 - 1)$

I don't know how to start.

Answer 1

The LHS is non-negative and $x_1^2+x_2^2\leq 1$, hence the inequality is non-trivial only if $y_1^2+y_2^2\leq 1$.

So we may assume that $X=(x_1,x_2)$ and $Y=(y_1,y_2)$ are two points inside the unit circle and the angle between them is $\theta$, then prove: $$ \left(\|X\|\|Y\|\cos\theta-1\right)^2\geq (1-\|X\|^2)(1-\|Y\|^2) \tag{1}$$ that is equivalent to: $$ \|X\|^2+\|Y\|^2 - 2\|X\|\|Y\|\cos\theta \geq \|X\|^2 \|Y\|^2\sin^2\theta\tag{2}$$ or to: $$ \|X-Y\|\geq \|X\|\|Y\|\left|\sin\theta\right|= 2[OXY]\tag{3} $$ that is trivial as soon as the projection of the origin $O$ on the $XY$-line still lies inside the unit circle.

Answer 2

Let us put $d=LHS-RHS$. View $d$ as a quadratic form in $y_1$ and $y_2$. Writing $d$ as a signed sum of squares using standard techniques, we obtain

$$ d=\frac{\bigg(y_2(1-x_1^2)-x_2(1-x_1y_1)\bigg)^2+(1-x_1^2-x_2^2)(y_1-x_1)^2}{1-x_1^2} $$ which is indeed nonnegative (note that when $1-x_1^2=0$, the RHS is zero).