MHB Checking if an interval value satisfies an expression

[Siron]

Hello!

Suppose a quartic polynomial with real roots where all its roots lie in the interval
$$x \in \left[\frac{5}{8}a-\frac{3}{8}\sqrt{15a^2-16b}, \frac{5}{8}a+\frac{3}{8}\sqrt{15a^2-16b}\right]$$
where $a \in \mathbb{R}$ and $b<(15/16)a^2$. Is there a way to check that at least one of those roots satisfies the inequality:
$$\frac{5ax-4x^2-b}{15ax-20x^2-3b}<0.$$

I solved the above inequality with wolfram alpha which displays the following solutions (after summarizing cases and taking $b<(15/16)a^2$ into account):
$$x \in \left[\frac{5a}{8}-\frac{1}{8}\sqrt{25a^2-16b}, \frac{3a}{8}-\frac{1}{8}\sqrt{\frac{3}{5}}\sqrt{15a^2-16b}\right]$$
or
$$x \in \left[\frac{3a}{8}+\frac{1}{8}\sqrt{\frac{3}{5}}\sqrt{15a^2-16b}, \frac{5a}{8}+\frac{1}{8}\sqrt{25a^2-16b} \right].$$

So ... in fact I think I only need to check wether the interval for my roots lies in one of the above intervals. However, I think this will result in too strong restrains on $b$. Is there a better alternative to solve the question?

Many thanks!
Kind regards,
Siron

 

[Siron]

I now see that my question is not that clear so I will give more details. The quartic polynomial which I refer to in my previous post is given by:
$$f(x) = 16x^4-40ax^3+(15a^2+24b)x^2-18abx+3b^2$$
where $a,b \in \mathbb{R}$. If the discriminant is positive and $b<(15/16)a^2$ then the quartic has four real roots (it's not difficult to derive that). So I'm wondering if its possible to prove that at least one of these real roots satisfies the inequality:
$$\frac{5ax-4x^2-b}{15ax-20x^2-3b}<0.$$
or equivalently lies in one of the two intervals:
$$x \in \left[\frac{5a}{8}-\frac{1}{8}\sqrt{25a^2-16b}, \frac{3a}{8}-\frac{1}{8}\sqrt{\frac{3}{5}}\sqrt{15a^2-16b}\right] := [l_1,r_1] $$
or
$$x \in \left[\frac{3a}{8}+\frac{1}{8}\sqrt{\frac{3}{5}}\sqrt{15a^2-16b}, \frac{5a}{8}+\frac{1}{8}\sqrt{25a^2-16b} \right] := [l_2,r_2].$$

I first thought about the intermediate value theorem and was thinking if I maybe could force $f(l_1)<0$ and $f(r_1)>0$ (or $f(l_2)<0$ and $f(r_2)>0$). However this gives very complicated expressions. Next, I tried to use the Samuelson-Laguerre inequality. Basically, this inequality implies an interval for the roots of the quartic. Applying the inequality, the four real roots of the quartic have to lie in the interval:
$$x \in \left[\frac{5}{8}a-\frac{3}{8}\sqrt{15a^2-16b}, \frac{5}{8}a+\frac{3}{8}\sqrt{15a^2-16b}\right]$$
and this was in fact my question/idea of the first post. Maybe I can force conditions such that the above interval lies within $[l_1,r_1]$ or $[l_2,r_2]$ but this gives very strong constraints on $b$.

Therefore, I'm wondering if there is another more efficient idea/method ... ?

Thanks!