Solving three inequalities with three variables

Question

I wish to show that the following three inequalities are impossible to satisfy, under the constraint that one and only one variable is allowed to be negative. The inequalities are: \begin{align} \ adf & < 0 \tag{1}\\ \ a + d + f & < 0 \tag{2}\\ \ ad + af + df & > 0 \tag{3}\\ \end{align} where $a<0$ and $d,f>0$. The variables are all real numbers ($a, d, f \in \mathbb R$).

I tried doing it by isolating $a$ in $(3)$ and then inserting $(2)$ after also isolating $a$ here. So from $$a > \dfrac {-df}{d+f}$$ and $$a < -(d+f),$$ which gave me $$-(d + f) > \dfrac {-df}{d+f},$$ then multiplying with $d+f$ gave $$-( d^2 + f^2 +2df) > -df,$$ which results in $$d^2 + f^2 + df < 0.$$ The last line is impossible, so it must be impossible to have one of the three variables negative and the other two positive. Are the method and the steps correct?

Edit: We also have the inequality $$2adf + a^2(d+f) + d^2(a+f) + f^2(a+d) < 0.$$

Answer 1

Yes, your solution is correct.

Alternatively, from $(3)$: $$ad+af+df>0 \iff a+d>-\frac{ad}{f}$$ Add $f$ to both sides: $$a+d+f>-\frac{ad}{f}+f$$ Consider this with $(2)$: $$0>a+d+f>-\frac{ad}{f}+f \Rightarrow 0>-\frac{ad}{f}+f \Rightarrow f^2-ad<0 \Rightarrow \\ \underbrace{f^2}_{>0}+\underbrace{(-ad)}_{>0}<0.$$ The contradiction shows that the the system of inequalities does not have a solution for the given constraints.

Answer 2

Your proof looks correct. Strictly speaking, you should start with the assumption that $a < 0$ and $d, f \ge 0$, but $(1)$ then implies that $d, f > 0$.

An alternative is to note that the expressions $$ a+d+f \, , \quad ad + af + df \, , \quad adf $$ are the “elementary symmetric polynomials” of $a, d, f$, and that suggests to consider the polynomial $$ p(x) = (x-a)(x-d)(x-f) = x^3 - (a+d+f)x^2 + (ad + af + df) x - adf \, . $$ The given inequalities imply that $p$ has only positive coefficients, so that all real roots must be negative. It follows that $a, d, f$ are all negative.