Perpendicular Vectors on Unit Sphere induce bound on components Working through an old preliminary exam.
Let $x=(x_1,...,x_n)^{T}$ and $y=(y_1,...,y_n)^{T}$ be column vectors in $\mathbb{R}^n$ and suppose $n\geq2$. Suppose also that $\vert\vert x\vert\vert=\vert\vert y\vert\vert=1$ and that $x\perp y$. Prove that
$x_1^2+y_1^2\leq 1$
If I remember correctly, on the exam I stated that because the norm and orthogonality are both inherited from the inner product just take an isometry that maps one of the vectors to $e_1$. The desired inequality follows easily.
However, I'm trying to now actually work through this without such appeals. I've deduced that the subscript 1 is in fact arbitrary, and that
$x_i^2+y_i^2\leq 1$
must hold for all $n$. I can't seem to get a componentwise estimate. If there is such thing as a hint without giving a full answer, I'd appreciate that first. Thank you.
 A: If $x_1 = 0$, the claim is straightforward. So assume $x_1 \ne 0$. We have
$$x \perp y \iff x_1y_1 + \cdots + x_ny_n = 0 \iff  y_1 = \frac{-(x_2y_2+ \cdots + x_ny_n)}{x_1} \tag 1$$
and
$$(x_2y_2 + \cdots + x_ny_n)^2 \le (y_2^2 + \cdots + y_n^2) (x_2^2 + \cdots + x_n^2) \tag 2$$
by the Schwarz inequality,  and
$$y_1^2 + \cdots +y_n^2 = x_1^2 + \cdots + x_n^2 \tag 3$$
since $\|x\| = \|y\| = 1$.
Substituting $(1)$ into $(3)$ and then using $(2)$ yields
$$ \frac{(y_2^2 + \cdots + y_n^2) (x_2^2 + \cdots + x_n^2)}{x_1^2} + y_2^2 + \cdots + y_n^2 \ge  x_1^2 + \cdots + x_n^2 $$
$$\iff (y_2^2 + \cdots + y_n^2) \left(\frac{x_2^2 + \cdots + x_n^2}{x_1^2} + 1 \right)\ge  x_1^2 + \cdots + x_n^2$$
$$\iff (y_2^2 + \cdots + y_n^2) \ge  \frac{1}{\frac{x_2^2 + \cdots + x_n^2}{x_1^2} + 1}(x_1^2 + \cdots + x_n^2)$$
$$\iff (y_2^2 + \cdots + y_n^2) \ge  \frac{x_1^2}{x_1^2 + \cdots + x_n^2}(x_1^2 + \cdots + x_n^2)$$
$$\iff (y_2^2 + \cdots + y_n^2) \ge  x_1^2$$
This gives the desired inequality:
$$y_1^2 + x_1^2 \le y_1^2 + (y_2^2 + \cdots + y_n^2) = 1$$
$\square$
