Again, a construction
Given a triangle $ABC$ with all its angles smaller than $90^\circ$, using standard notations ($\tau$ is the semiperimetr, $r$ the radius of the incircle.), prove that : $\displaystyle \sum_{cyc} [(\tan A) ^3 \tan B ] \geq [\frac{2\tau r}{\tau^2-25r^2}]^2$
I post here my solution, hoping for more.
Equivalently, we have to show that $\displaystyle \sum_{cyc} \frac{(\tan A)^2}{\tan B} \geq [\frac{2\tau r}{\tau^2-25r^2}]\cot A \cot B \cot C $.
From Cauchy-Schwarz, it suffices to show that $\displaystyle \sum_{cyc} \tan A \geq [\frac{2\tau r}{\tau^2-25r^2}]\cot A \cot B \cot C$ .
Since, $\cot A \cot B \cot C=\frac{1}{\tan A \tan B \tan C}$ and $\displaystyle \sum_{cyc} \tan A=\tan A \tan B \tan C$ we just have to prove that $\tan A \tan B \tan C \geq \frac{2\tau r}{\tau^2-25r^2}$.
Finally, it's fairly well-known that $\sin A \sin B \sin C=\frac{\tau r}{2R^2}$ and $\cos A \cos B \cos C=\frac{\tau^2-(2R+r)^2}{4R^2}$ where $R$ is the radius of the circumcircle.
So, it suffices to show that $\frac{2\tau r}{\tau^2-(2R+r)^2} \geq \frac{2\tau r}{\tau^2-25r^2} \Rightarrow R \geq 2r$, which is the well-known Euler inequality.
Let's see different solutions to this problem, or maybe proofs of the identities used above for $\cos A \cos B \cos C, \sin A \sin B \sin C$