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Now, this question, though is straightforward in nature is making me weary and I am unable to find a way to obtain a nice solution set for it.

Let's say, given that I have a inequality,

$$ \Big(g-12 a (a Q^{2}-h u)\Big)^{3}<16 \Big(g^{3}+36 a Q^{2} g-18 a h u g) +54 a^{2}(u^{2}-h^{2} Q^{2})\Big)^2$$

I want to find the expression of $a$ in terms of $u,Q,g,h$ which satisfies the given condition.

My take on it was to expand all the terms and solve for the roots obtaining a univariate polynomial in $a$, but I observed that it became an expression like this:

$$110592 a^6 Q^6+331776 a^5 h Q^4 u+11664 a^4 h^4 Q^4-27648 a^4 h^2 Q^4+308448 a^4 h^2 Q^2 u^2+27648 a^4 Q^4+11664 a^4 u^4+7776 a^3 h^5 Q^2 u-15552 a^3 h^4 Q^4-63072 a^3 h^3 Q^2 u+102816 a^3 h^3 u^3+15552 a^3 h^2 Q^4+15552 a^3 h^2 Q^2 u^2+55296 a^3 h Q^2 u+7776 a^3 h u^3-15552 a^3 Q^2 u^2-432 a^2 h^8 Q^2+1296 a^2 h^6 Q^2+1728 a^2 h^6 u^2-5184 a^2 h^5 Q^2 u+5184 a^2 h^4 Q^4+1008 a^2 h^4 Q^2-31536 a^2 h^4 u^2+10368 a^2 h^3 Q^2 u-10368 a^2 h^2 Q^4-4176 a^2 h^2 Q^2+30240 a^2 h^2 u^2-5184 a^2 h Q^2 u+5184 a^2 Q^4+2304 a^2 Q^2-432 a^2 u^2-144 a h^9 u+288 a h^8 Q^2+576 a h^7 u-1152 a h^6 Q^2+1440 a h^5 u+1728 a h^4 Q^2-4032 a h^3 u-1152 a h^2 Q^2+2160 a h u+288 a Q^2+4 h^{12}-24 h^{10}+60 h^8-144 h^6+252 h^4-216 h^2+68 $$

Which is an equation of order $6$ and hence no proper symbolic results can be obtained even using software's like Mathematica.

I want to know the values that $a$ can take, any method that I am missing here to do so?

Thanks.

mnuizhre
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    As you say , it is unlikely that there is a closed form for $a$. So, I think, we must know all values except $u$ , and then we just have a univariate polynomial. – Peter Mar 22 '22 at 15:16
  • I think I didn't make it clear in the question but this IS a univariate polynomial in $a$, others are just constants. – mnuizhre Mar 22 '22 at 15:16
  • @Peter $u,Q,g,h$ are constants. – mnuizhre Mar 22 '22 at 15:18
  • Finding roots is only "straight forward" , if there are no parameters. Even the case of degree $3$ which can actually be expressed as a closed form (with parameters) is so complicated that it does not have a merit to use it. – Peter Mar 22 '22 at 15:19
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    Note that it's not true that there are no explicit symbolic solutions. See for example this post. But it is true that there is no general expression in radicals (the Abel-Ruffini theorem). For any fixed set of constants, it is straightforward to numerically approximate the roots. Mathematica can do this. In neighborhoods of any fixed set, it would be possible to investigate the behavior using the implicit function theorem (roots are implicitly functions of the coefficients of the polynomial), but this is complicated. – davidlowryduda Mar 22 '22 at 15:20
  • So, there isn't any way that I can get a bound for $a$ for this expression? (I mean, not necessarily finding analytical roots) – mnuizhre Mar 22 '22 at 15:21

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