<s>[INST] How prove this inequality $\sum\tan{\frac{A}{2}}-\frac{\sqrt{3}-1}{8}\prod\csc{\frac{A}{2}}\le 1$ In $\Delta ABC$,show that
$$\tan{\frac{A}{2}}+\tan{\frac{B}{2}}+\tan{\frac{C}{2}}-\frac{\sqrt{3}-1}{8}\csc{\frac{A}{2}}\csc{\frac{B}{2}}\csc{\frac{C}{2}}\le 1$$
I tried also $$\left(\tan{\frac{A}{2}}+\tan{\frac{B}{2}}+\tan{\frac{C}{2}}\right)^2\ge 3\left(\tan{\frac{A}{2}}\tan{\frac{B}{2}}+\tan{\frac{A}{2}}\tan{\frac{C}{2}}+\tan{\frac{C}{2}}\tan{\frac{B}{2}}\right)=3,$$ but we get there something, which impossible to kill during a competition.
 [/INST] Easy to show that $\sum\limits_{cyc}\tan\frac{\alpha}{2}=\frac{4R+r}{p}$ and $\prod\limits_{cyc}\sin\frac{\alpha}{2}=\frac{r}{4R}$.
Also, if $M$ is a center gravity of the triangle and $I$ is a center of the inscribed circle in the triangle, 
so $MI^2=\frac{p^2+5r^2-16Rr}{9}$, which gives $p\geq\sqrt{16Rr-5r^2}$.
Let $R=2xr$. Hence, we need to prove that
$$\frac{4R+r}{p}-\frac{(\sqrt3-1)R}{2r}\leq1$$ or
$$p\geq\frac{2r(4R+r)}{2r+(\sqrt3-1))R},$$
for which it's enough to prove that
$$\sqrt{16Rr-5r^2}\geq\frac{2r(4R+r)}{2r+(\sqrt3-1))R}$$ or
$$(32x-5)((\sqrt3-1)x+1)^2\geq(8x+1)^2$$ or
$$(x-1)(32(2-\sqrt3)x^2-5(2-\sqrt3)x+3)\geq0,$$
which is obviously true for $x\geq1$.
Done!
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