Let $a,b,c$ be any three complex numbers lying on or inside the unit circle $|z|= 1.$ Let $|a-b|=r, |b-c|=s, |c-a|=t, $ with $r\leq t,$ then the inequality $$r^2s^2+s^2\leq 2r^2+2t^2$$ seems to be true. Whether the circumradius of the triangle with vertices $a,b,c$ will help us in proving this? or is there any other alternative approach?
Asked
Active
Viewed 110 times
2
-
A new hypothesis part $r\leq t$ is added. – user159888 Dec 03 '17 at 16:23
-
This is still wrong. See my answer. – stressed out Dec 03 '17 at 19:09
-
@user159888 Please, if you are ok, you can accept the answer and set it as solved. Thanks! – user Jan 17 '18 at 23:00
2 Answers
2
The problem, even with the assumption that $ r \leq t$, is again wrong.
Take $a=(-1,0)$, $b=(0,1)$ and $c=(0,-1)$.
Then $r=t=\sqrt{2}$ and $s=2$
$$r^2s^2+s^2=2\times 4+4=12$$ $$2r^2+2t^2=2\times2+2\times2=8$$
Also, note that since the module of a complex number is continuous with respect to its coordinates and your inequality is a polynomial in $r,s$ and $t$, even the stronger assumption $r < t$ won't work. For example, you can slightly change $c$ to make $r<t$, while the value of the polynomial $r^2s^2+s^2-2(r^2+t^2)$ won't change much because of continuity. Therefore, the inequality will still remain wrong.
For example, modify $c$ slightly and take it to be $c=(\frac{1}{n},-1)$ for some large $n$.
stressed out
- 8,130
-
-
Thank you very much. I am still thinking about what kind the stricter conditions could be imposed on r, s, t to get the inequality!? – user159888 Dec 04 '17 at 02:00
-
@user159888: well, first tell me why do you think that there must be an inequality of this form at all? Where does your interest in that expression come from? – stressed out Dec 04 '17 at 03:50
1
consider a, b, c on the circle with:
$r,s \to 2$ and $t\to 0$
thus
$r^2s^2+s^2 = 20 \geq 2r^2+2t^2=8$
user
- 154,566
-
-
Thank you. As it is, not true. But when $r\leq t$ it seems to be true. I added in the hypothesis part. – user159888 Dec 03 '17 at 16:19
-