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
$$\frac{\sqrt{3}}{2}\left(\cos\frac{B-C}{2}+\cos\frac{C-A}{2}+\cos\dfrac{A-B}{2}\right)=\frac{\sqrt{3}}{2}\sum\cos\frac{A}{2}\cos\frac{B}{2}+\sin\frac{B}{2} \sin\frac{C}{2}$$
 A: $$2\sum \sin \frac{A}{2}\sin \frac{B}{2}\leq  \frac{2}{3}\left(\sum \sin \frac{A}{2}\right)^2\leq\sum \sin \frac{A}{2}$$
$$=\sum \cos \frac{A}{2}\cos \frac{B}{2}-\sum \sin \frac{A}{2}\sin \frac{B}{2},$$
$$  \frac{\sqrt{3}}{2}\sum \cos \frac{A-B}{2}=\frac{\sqrt{3}}{2} \left(\sum \cos \frac{A}{2}\cos \frac{B}{2}+\sum \sin \frac{A}{2}\sin \frac{B}{2}\right)$$
$$\leq \frac{2\sqrt{3}}{3} \sum \cos \frac{A}{2}\cos \frac{B}{2}\leq \frac{2\sqrt{3}}{9}\left (\sum \cos \frac{A}{2}\right)^2\leq \sum \cos \frac{A}{2}.$$
A: Denote the difference $\text{LHS}-\text{RHS}$ by $S(A,B,C)$. We need to show that $S(A,B,C)\ge0$. We begin with a simple

Proposition. For $\alpha\in[0,\pi/2]$ we have the following inequality
  $$2\cos\frac{\pi-\alpha}4-\sqrt3\cos\frac{\pi-3\alpha}4<\sqrt3$$
  Proof. $\frac{\pi-3\alpha}4\in[-\frac\pi8,\frac\pi4]$, thus
  $$2\cos\frac{\pi-\alpha}4-\sqrt3\cos\frac{\pi-3\alpha}4\le2-\sqrt3\cos\frac\pi4<\sqrt3.\quad\square$$

Now suppose $A\le\pi/2$ (such an angle always exists in a triangle; e.g. take $A$ to be the smallest one among $A,B,C$). Reformulate the desired inequality as
$$\cos\frac A2+2\cos\frac{\pi-A}4\cos\frac{B-C}4\ge\sqrt3\cos^2\frac{B-C}4-\frac{\sqrt3}2+\sqrt3\cos\frac{\pi-3A}4\cos\frac{B-C}4$$
Let $t=\cos\frac{B-C}4\in(0,1]$. Consider the function
$$f(x)=-\sqrt3t^2+(2\cos\frac{\pi-A}4-\sqrt3\cos\frac{\pi-3A}4)x$$
By the proposition at the beginning of this proof, The axis of symmetry of $f$ is
$$x=\frac{2\cos\frac{\pi-A}4-\sqrt3\cos\frac{\pi-3A}4}{2\sqrt3}<\frac12$$
And since $-\sqrt3<0$, the parabola opens downwards. Consequently $f(t)\ge f(1)$. Furthermore, $f(t)>f(1)$ if $t\not=1$. In other words, $S(A,B,C)\ge S(A,\frac{B+C}2,\frac{B+C}2)$ with the equality if and only if $B=C$.  
We're almost done. Observe that $S$ is continuous in $A, B, C$. We may extend the range of $A,B,C$ to $[0,\pi]^3\cap\{A+B+C=\pi\}$ without destroying the above argument. This is a compact set, on which $S$ attains its minimum $S_0$ at the point $(A_0, B_0, C_0)$. WLOG we may assume $A\le\pi/3$.  If $B_0\not=C_0$, then $S_0>S(A_0,\frac{B_0+C_0}2,\frac{B_0+C_0}2)$, contradicting the minimality of $S_0$. Thus $B_0=C_0$. But $B_0+C_0=\pi-A_0\le\pi$, so $B_0\le\pi/2$. Applying the above argument once again, we see $A_0=C_0$. Now we conclude that $A_0=B_0=C_0=\pi/3$, hence $S_0=S(\pi/3,\pi/3,\pi/3)=0$. This shows that $S\ge S_0=0$.
A: 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!
