How can I show that the following equation is always negative?

Question

I have the following function,

$$f(a,a_p) = -2 a_p + a (-2 + a + a_p + 3 a a_p + (1 + 2 (-2 + a) a) a_p^2)$$ and the following condition $$1>a_p>a>0$$ I used the FindInstance function in Mathematica and did not find any instance of $f(a,a_p)$ being positive. So, I know that $f(a,a_p)$ is always negative. However, I am trying to prove this analytically.

Here is what I tried: I collected all the $a$ terms and have the following cubic equation, $$ 2 a^3 a_p^2 + a^2 (1 + 3 a_p - 4 a_p^2) - a (2 - a_p - a_p^2) - 2 a_p$$ where all the coefficients of $a,~a^2,~\text{and}~a^3$ are positive. But I am stuck here trying to analyze the cubic equation.

Thanks in advance.

Answer 1

Fixing $a_p,$ consider your cubic $g(a).$ We have $g(0)=-a_p$ and $$g(1)=2a_p^2+1+3a_p-4a_p^2-2+a_p+a_p^2-2a_p=-a_p^2+2a_p-1=-(a_p-1)^2<0.$$ Since $\lim_{a\to \infty} g(a)=+\infty,$ we see $g$ has a zero on the interval $(1,\infty).$ By Descartes' rule of signs, $g$ has at most one positive zero, so it can't have any zero on $(0,1),$ and thus it's always negative there as MathLab suggests.