Solution for three variable Simultaneous Equations

Question

I have two Equations : $$1)\; abc=1$$ $$2)\; a+b+c=1$$

And the constraint that $a,b,c$ are positive real numbers. I have to prove that there exist no solution for the given constraint.

My attempt: I substituted $a$ in terms of$ b$ and $c$ in equation 2 It resulted in a Diophantine equation. I tried getting constraint for value of $b$ by using the inequality that discriminant of a quadratic must be positive for a real root to exist but the inequality results in a 4th degree equation in $c$.

Answer 1

AM-GM will help you: $$\frac{a+b+c}{3}\geq \sqrt[3]{abc}$$ which gives $$1\geq 3$$

Answer 2

Since we have inequality between arithmetic and geometric mean

$$1= a+b+c\geq 3\sqrt[3]{xyz} =3$$

we have no solution.