Prove: $k^3 - k( b c + c a + a b ) + 2 a b c = 0$ always has a negative root with all positive parameters $a, b, c$

Question

Prove: $k^3 - k( b c + c a + a b ) + 2 a b c = 0$ always has a negative root with all positive parameters $a, b, c$

I tried: Write $f(x)=x^3-x(ab+ac+bc)+2abc$ then $f(-\infty)=-\infty,f(0)>0$. Now use the Intermediate Value Theorem. I can' t continue. Help me! Thanks!

Answer 1

Since $a,b,c$ are positive the value of $f(0)=2abc>0$. And $f(-\infty)=-\infty$. So by intermediate value property, there exists $c\in(-\infty,0)$ such that $f(c)=0$

Intermediate Value Theorem If there is a continuous function $f:[a,b]\rightarrow \mathbb{R}$. There is some $L$ such that $f(a)<L<f(b)$ or $f(a)>L>f(b)$, then there exists a point $c \in [a,b]$, such that $f(c)=L$.

Look at the theorem and see if you can match what $a,b,L,c$ are in our case.

Answer 2

Alt. hint (without IVT):   by Vieta's relations the product of the three roots is $\,-2abc \lt 0\,$. Since the polynomial has real coefficients:

• either all three roots are real, in which case at least one must be negative;

• or one root is real and the other two are complex conjugates, in which case the product of the two complex roots equals the square of their modulus, so the real root must be negative.

Answer 3

Additionally to the already observed $f(0)=2abc>0$ one also has $$ f(\sqrt[3]{abc})=-\sqrt[3]{abc}(ab+bc+ac)+3abc $$ As per mean inequalities $$ \frac{ab+bc+ac}3\ge\sqrt[3]{a^2b^3c^2} $$ this function value is non-positive, which implies that there are 2 positive real roots or a double root at $\sqrt[3]{abc}$ in addition to the negative root.