<s>[INST] Prove this trigonometric inequality about the angles of $\triangle ABC$ 
In $\Delta ABC$ show that
  $$\cos{\frac{A}{2}}+\cos\frac{B}{2}+\cos\frac{C}{2}\ge \frac{\sqrt{3}}{2} \left(\cos\frac{B-C}{2}+\cos\frac{C-A}{2}+\cos\frac{A-B}{2}\right)$$

since
$$\frac{\sqrt{3}}{2}\left(\cos\frac{B-C}{2}+\cos\frac{C-A}{2}+\cos\dfrac{A-B}{2}\right)=\frac{\sqrt{3}}{2}\sum\cos\frac{A}{2}\cos\frac{B}{2}+\sin\frac{B}{2} \sin\frac{C}{2}$$
 [/INST] We can make also the following thing.
We'll prove that for all triangle the following inequality is true.
$$\sin\alpha+\sin\beta+\sin\gamma\geq\frac{\sqrt{3}}{2}(\cos(\alpha-\beta)+\cos(\alpha-\gamma)+\cos(\beta-\gamma)).$$
Indeed, let $R$ be a radius of  circumcircle, $r$ be a radius of inscribed circle and $p$ be a semiperimeter of the triangle.
Hence, $$\sum_{cyc}\sin\alpha=\sum_{cyc}\frac{a}{2R}=\frac{p}{R}.$$
In another hand, easy to show that
$$\sum_{cyc}\cos(\alpha-\beta)=\frac{p^2-2R^2+2Rr+r^2}{2R^2}.$$
Thus, we need to prove that
$$\frac{p}{R}\geq\frac{\sqrt3(p^2-2R^2+2Rr+r^2)}{4R^2}$$ or
$$\sqrt3p^2-4Rp-2\sqrt3R^2+2\sqrt3Rr+\sqrt3r^2\leq0$$ and since
$$-2\sqrt3R^2+2\sqrt3Rr+\sqrt3r^2=-2\sqrt3R(R-r)+\sqrt3r^2\leq-2\sqrt3\cdot2r(2r-r)+\sqrt3r^2<0,$$
we need to prove that
$$p\leq\frac{2R+\sqrt{10R^2-6Rr-3r^2}}{\sqrt3}.$$
Now, by Gerretsen inequality $p\leq\sqrt{4R^2+4Rr+3r^2}$.
Id est, it's enough to prove that
$$\sqrt{4R^2+4Rr+3r^2}\leq\frac{2R+\sqrt{10R^2-6Rr-3r^2}}{\sqrt3}.$$
Let $R=2rx$.
Hence, $x\geq1$ and  we need to prove that:
$$\sqrt{16x^2+8x+3}\leq\frac{4x+\sqrt{40x^2-12x-3}}{\sqrt3}$$ or
$$2x^2-9x-3+2x\sqrt{40x^2-12x-3}\geq0$$ or
$$2x^2+x-3+2x\left(\sqrt{40x^2-12x-3}-5\right)\geq0$$ or
$$(x-1)\left(2x+3+\frac{8x(10x+7)}{\sqrt{40x^2-12x-3}+5}\right)\geq0,$$
which is obvious.
Now, since the inequality
$$\sin\alpha+\sin\beta+\sin\gamma\geq\frac{\sqrt{3}}{2}(\cos(\alpha-\beta)+\cos(\alpha-\gamma)+\cos(\beta-\gamma))$$
is true for all triangle, this inequality is true also for any acute-angled triangle 
and after replacing in the last inequality 
$\alpha$ on $90^{\circ}-\frac{\alpha}{2}$, $\beta$ on $90^{\circ}-\frac{\beta}{2}$ and $\gamma$ on $90^{\circ}-\frac{\gamma}{2}$ we'll get the starting inequality.
Done!
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