<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] 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 $S\ge S_0=0$.
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