Find all $ a,b,c \in \mathbb{R} $ that satisfies the equations, $ \bullet $ $ a+b+c=63$ and $ \bullet $ $ab+bc+ac=2021$? Find all $ a,b,c \in \mathbb{R} $ that satisfies the equations,
$ \bullet $ $ a+b+c=63$
$ \bullet $ $ab+bc+ac=2021$ ?
I try to solve this problem but going to the result that this problem has no solutions at all ... My attempt about the solution is that which is best of other ones..
$ \forall a,b,c \in \mathbb{R} $, we have
$(a+b+c)^2 = a^2 + b^2 + c^2 +2ab +2bc + 2ca $
from here we get,
$a^2 + b^2 + c^2= (a + b + c)^2 — 2(ab + bc + ca)$
$a^2 + b^2 + c^2= 63^2 — 2(2021)$
$= 3969 — 4042$
$= -73$
which is impossible since $a^2 + b^2 + c^2 \geq 0$.
Thus there are no such $ a, b, c \in \mathbb{R} $ that satisfy the given equation
Is There any set of Real numbers satisfy the conditions?

By Vieta's formula we have
$$f(x) = x^3 - 63x^2 + 2021x -abc = 0$$
This function has only one real root since $f'(x) > 0$ for all $x$.
