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Let $t \in \left[\frac{\sqrt{3}}{3};1 \right)$. Prove that

$$\frac{{27{t^6}\left( {4{t^4} + 3{t^3} + 16{t^2} + 3t + 4} \right)}}{{(2t + 1)\left( {4{t^6} - 21{t^5} + 36{t^4} + 7{t^3} - 24{t^2} + 4} \right)}} \geq 3\left( {35\sqrt 3 - 48} \right)$$ In a problem I'm working on, the following issue arises:

When I attempt to take the derivative, the function on the left-hand side is not monotonic, and the calculation is quite involved. Using computational software, the inequality holds true. Equality occurs when $t = \frac{1}{\sqrt{3}}$. I'm seeking a more intelligent approach to solve this. Thank you.

Gary
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Math_fun2006
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    $D(t)$ is $>0$, hence $N(t)/D(t)\ge C$ is equivalent to $N(t)-CD(t)\ge0$. – Anne Bauval Mar 23 '24 at 06:41
  • @AnneBauval Continue with this approach. I've also attempted it, and the resulting polynomial is of degree 10. – Math_fun2006 Mar 23 '24 at 06:44
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    Isn't there anything in the problem before that would allow us to simplify the expression as trigonometric (i.e. $t=\tan \theta$ with $\theta\in[\frac{\pi}6,\frac{\pi}4)$) rather than this ugly high degree polynomial ? – zwim Mar 23 '24 at 10:29

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