If $a^2+b^2+c^2=1$ and $x^2+y^2+z^2=1$, can it be shown that $(ax)^2+(by)^2+(cz)^2=1$?

Question

If $a^2+b^2+c^2=1$ and $x^2+y^2+z^2=1$, can it be shown that $(ax)^2+(by)^2+(cz)^2=1$ for all real a, b, c, x, y, z?

This statement doesn't seem true. Even $(a, b, c) = (0, 0, 1)$ and $(x, y, z) = (1, 0, 0)$ doesn't work. — Elliot Yu, Dec 17 '20 at 10:18
You mean, like when all variables are evaluated as $1/\sqrt3$? — Oscar Lanzi, Dec 17 '20 at 10:19
Because, based on zwim's logic below, it is possible $\forall t$ when $a = x = 0$, $b = y = cos(t)$, $c = z = sin(t)$. But I can not show, that these are all the values for which the conditions are true. — Dávid Laczkó, Dec 17 '20 at 13:38

score 3 · Answer 1 · answered Dec 17 '20 at 10:30

Even in the reduced case with only two couples of variables $\begin{cases} c=z=0\\a=y=\cos(t)\\b=x=\sin(t)\end{cases}$

Then $(ax)^2+(by)^2+(cz)^2=2\big(\cos(t)\sin(t)\big)^2=\frac 12\sin(2t)^2\le \frac 12\neq 1$

So there are infinity of values for which the equation is not verified.

score 2 · Answer 2 · answered Dec 17 '20 at 10:18

2

No. Take $a=c=y=z=0$ and $b=x=1$.

answered Dec 17 '20 at 10:18

lhf

2 Answers2