$\frac1{12}(\sin^2A+\sin^2B+\sin^2C)\leq \sin^2\frac12B\sin^2\frac12C+\sin^2\frac12C\sin^2\frac12A+\sin^2\frac12A\sin^2\frac12B$ for $A+B+C=180^\circ$

For any $(A,B,C)$ satisfying $A+B+C=180^\circ$, the inequality holds.

Proof :

Letting $s:=\tan\dfrac A2$ and $t:=\tan\dfrac B2$, one can write $$\sin A=\frac{2s}{1+s^2},\ \cos A=\frac{1-s^2}{1+s^2},\ \sin B=\frac{2t}{1+t^2},\ \cos B=\frac{1-t^2}{1+t^2}\tag1$$

In the following, using $$C=180^\circ-A-B\tag2$$ let us represent $\text{(RHS) $-$ (LHS)}$ by $s,t$.

Using $(1)(2)$, one has $$\begin{align}&\text{(RHS) $-$ (LHS)} \\\\&=\frac{1-\cos A}{2}\cdot\frac{1-\cos B}{2}+\frac{1-\cos B}{2}\cdot\frac{1-\cos C}{2}+\frac{1-\cos C}{2}\cdot\frac{1-\cos A}{2} \\&\qquad\quad -\frac{1}{12}(\sin^2 A+\sin^2 B+\sin ^2C) \\\\&=\frac{(1-\cos A)(1-\cos B)}{4}+\frac{(1-\cos B)(1+\cos(A+B))}{4} \\&\qquad\quad +\frac{(1+\cos(A+B))(1-\cos A)}{4}-\frac{1}{12}\bigg(\sin^2 A+\sin^2 B+\sin ^2(A+B)\bigg) \\\\&=\frac{s^2t^2}{(1+s^2)(1+t^2)}+\frac{t^2(1-st)^2}{(1+s^2)(1+t^2)^2}+\frac{s^2(1-st)^2}{(1+s^2)^2(1+t^2)} \\&\qquad\quad -\frac{1}{12}\bigg(\frac{4s^2}{(1+s^2)^2}+\frac{4t^2}{(1+t^2)^2}+\frac{4(s+t)^2(1-st)^2}{(1+s^2)^2(1+t^2)^2}\bigg) \\\\&=\frac{P}{3(1+s^2)^2(1+t^2)^2}\end{align}$$

where $$P=9 s^4 t^4 + 4s^4 t^2 - 14 s^3 t^3 - 4 s^3 t + 4 s^2 t^4 + 9 s^2 t^2 + s^2 - 4 s t^3 + t^2 - 2 s t $$

Now, our problem is reduced to proving that $P$ is non-negative for any $s,t$.

Here, there may exist real numbers $a,b,c,d,e,f,g,h$ such that $$P=\bigg((ast+b)(cst+d)\bigg)^2+\bigg((es+ft)(gst+h)\bigg)^2$$ where $a,c,e,g$ are non-negative real numbers.

Comparing the coefficients, one gets $$\begin{cases}9=a^2c^2 \\4= e^2 g^2 \\-14=2a^2cd+ 2 a b c^2 + 2 e f g^2 \\-4=2 e^2 g h \\4=f^2 g^2 \\9=a^2 d^2+ 4 a b c d + b^2 c^2+ 4 e f g h \\1=e^2 h^2 \\-4= 2 f^2 g h \\1= f^2 h^2 \\-2= 2 a b d^2 + 2 b^2 c d + 2 e f h^2 \\0=b^2 d^2 \end{cases}$$ which can be simplified as $$\begin{cases}3=ac \\2=eg \\-1=ad+bc \\-1=eh \\-2= f g \\1= f h \\0=bd \end{cases}$$

Taking $b=0$ and $e=1$, one has $h=-1,g=2,f=-1$ and $$\begin{cases}3=ac \\-1=ad \end{cases}$$ So, one can see that $(a,c,d)=(1,3,-1)$ works.

It follows from these that $$P=\bigg(st(3st-1)\bigg)^2+\bigg((s-t)(2st-1)\bigg)^2\geqslant 0$$

Therefore, one finally has $$(\text{RHS})-(\text{LHS})=\frac{s^2t^2(3st-1)^2+(s-t)^2(2st-1)^2}{3(1+s^2)^2(1+t^2)^2}\geqslant 0.\ \quad\blacksquare$$

Added : Since $s=\tan\frac A2,t=\tan\frac B2$ are defined only when $\frac A2\not=90+180k_1$ and $\frac B2\not=90+180k_2$ where $k_1,k_2$ are integers, one also has to prove that if either $A=180+360k_1$ or $B=180+360k_2$, then the inequality holds.

• If $A=180+360k_1$, then since $C=-B-360k_1$, $$(\text{RHS})-(\text{LHS})=\bigg(\frac 43+\frac 53\sin^2\frac{B}{2}\bigg)\sin^2\frac B2\geqslant 0$$ whose equality holds if and only if $(A,B,C)=(180+360k_1,360k_3,-360k_3-360k_1)$ where $k_3$ is an integer.

• If $B=180+360k_2$, then since $C=-A-360k_2$, $$(\text{RHS})-(\text{LHS})=\bigg(\frac 43+\frac 53\sin^2\frac{A}{2}\bigg)\sin^2\frac A2\geqslant 0$$ whose equality holds if and only if $(A,B,C)=(360k_4,180+360k_2,-360k_4-360k_2)$ where $k_4$ is an integer.

Under the condition that $s,t$ are defined, the equality holds if and only if $$\small\begin{align}& st(3st-1)=0\quad \text{and}\quad (s-t)(2st-1)=0 \\\\&\iff st=s-t=0\quad\text{or}\quad 3st-1=s-t=0 \\\\&\iff (s,t)=(0,0),\bigg(\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}\bigg),\bigg(-\frac{1}{\sqrt 3},-\frac{1}{\sqrt 3}\bigg) \\\\&\iff \bigg(\tan\frac A2,\tan\frac B2\bigg)=(0,0),\bigg(\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}\bigg),\bigg(-\frac{1}{\sqrt 3},-\frac{1}{\sqrt 3}\bigg) \\\\&\iff \bigg(\frac A2,\frac B2\bigg)=(180n_1,180n_2),(30+180n_3,30+180n_4),(150+180n_5,150+180n_6) \\\\&\iff (A,B)=(360n_1,360n_2),(60+360n_3,60+360n_4),(300+360n_5,300+360n_6)\end{align}$$ where $n_1,n_2,n_3,n_4,n_5,n_6$ are integers.

Finally, under the condition that $A,B,C$ are non-negative, the equality holds if and only if $$(A,B,C)=(60^\circ,60^\circ,60^\circ),(0^\circ,0^\circ,180^\circ),(0^\circ,180^\circ,0^\circ),(180^\circ,0^\circ,0^\circ).$$

This is only a special case

I managed to prove the inequality when $A,B,C$ are the angles of a triangle. I expect some kind of transformation to ensure that the general case $A+B+C = 180^\circ$ can be accommodated, but I acknowledge that I haven't investigated this.

If we do assume that $A,B,C$ are the angles of a triangle with side lengths $a,b,c$ respectively opposite $A,B,C$, then there are certain well-known formulas that tell us what the sine terms are, when expressed using $a,b,c$ and the quantity $s = \frac{a+b+c}{2}$ (the semi-perimeter).

These are (note : the cyclic counterparts of these equalities hold for $B,C$) : $$ \sin\frac A2 = \sqrt{\frac{(s-b)(s-c)}{bc}} \\ \sin A = \sqrt{\frac{4s(s-a)(s-b)(s-c)}{b^2c^2}} $$

They can be found on page 3 of this document that discusses trigonometric inequalities in general.

With this, we proceed to rewrite the inequality using the quantities $s,a,b,c$. Before doing this, we revert to a common inequality notation which is considered useful even for expressing inequalities. Suppose we have a function $f(x,y,z)$ of three arguments. Within the realm of this answer (and more generally as well), $\displaystyle\sum_{cyc} f(p,q,r)$ is defined as the quantity $f(p,q,r) + f(q,r,p)+f(r,p,q)$. It's the "cyclic" sum , if one likes.

With this, we can now write : $$ \frac{\sin^2 A + \sin^2 B + \sin^2 C}{12} = \sum_{cyc} \frac{s(s-a)(s-b)(s-c)}{3b^2c^2} $$

and $$ \sin^2 \frac{B}{2} \sin^2\frac A2 + \sin^2 \frac{C}{2} \sin^2\frac A2+\sin^2 \frac{B}{2} \sin^2\frac C2 \\ = \sum_{cyc} \frac{(s-a)^2(s-b)(s-c)}{a^2bc} $$

which means that we have to prove : $$ \sum_{cyc} \frac{s(s-a)(s-b)(s-c)}{3b^2c^2}\leq \sum_{cyc} \frac{(s-a)^2(s-b)(s-c)}{a^2bc} $$

Now, we want to get rid of the common denominator : for this, we multiply by $3a^2b^2c^2$ on both sides to get the equivalent inequality :$$ \sum_{cyc} a^2s(s-a)(s-b)(s-c) \leq \sum_{cyc} 3bc(s-a)^2(s-b)(s-c) $$

The term $(s-a)(s-b)(s-c)$ is common to each summation : it's a positive quantity, and dividing out by it, we get : $$ \sum_{cyc} a^2s \leq \sum_{cyc} 3bc(s-a) $$

We need one more transform now.

At this point, we use the Ravi substitution (named after Ravi Vakil). In short, we set $s-a = z, s-b = y , s-c = x$ . The idea of the substitution is the following : the condition that $a,b,c$ are the sides of a triangle with semi-perimeter $s$, is equivalent to the condition that $x,y,z$ are positive real numbers.

Under the substitution, we have $a=x+y,b=x+z,c=y+z$ , as can be easily checked. From now on, all cyclic inequalities will cycle over $x,y,z$ instead of $a,b,c$. So, we get : $$ \sum_{cyc} a^2 s = \sum_{cyc} (x+y)^2(x+y+z) $$

and $$ \sum_{cyc} 3bc(s-a) = \sum_{cyc} 3z(x+z)(y+z) $$

Thus, we come down to comparing : $$ \sum_{cyc} (x+y)^2(x+y+z) \leq \sum_{cyc} 3z(x+z)(y+z) $$

Now, cyclicity often hides a lot of terms that can be cancelled upon viewing for a short period. Let me clarify this by expanding the brackets on both sides : $$ \sum_{cyc} (x^3+y^3 +3x^2y+3xy^2+x^2z+y^2z+2xyz) \leq \sum_{cyc} (3z^3+3yz^2+3xz^2+3xyz) $$

Note that $\sum_{cyc} x^3 = \sum_{cyc} y^3 = \sum_{cyc} z^3$. Therefore, that part cancels to leave the equivalent $$ \sum_{cyc} (3x^2y+3xy^2+x^2z+y^2z+2xyz) \leq \sum_{cyc} (z^3+3yz^2+3xz^2+3xyz) $$

Now notice that there's a $2xyz$ on the left and $3xyz$ on the right. Resolving that:$$ \sum_{cyc} (3x^2y+3xy^2+x^2z+y^2z) \leq \sum_{cyc} (z^3+3yz^2+3xz^2+xyz) $$

Now observe, that $\sum_{cyc} x^2y = \sum_{cyc} z^2 x$, and observe that $\sum_{cyc} xy^2 = \sum_{cyc} yz^2$ (write this down if it's uncomfortable initially!). Thus, we are reduced to : $$ \boxed{\sum_{cyc} (x^2z+xz^2) \leq \sum_{cyc}{(z^3+3xyz)}} $$

This boxed inequality is true and well-known, see the $t=1$ special case here. It's the most typical application of the Schur inequality for three variables.

Having worked through the substitutions and manipulations in reversible fashion, we conclude that the original inequality is true as well. Furthermore, the equality case for the Schur-inequality is attained when $x=y=z$ (or some of them are zero,which is ruled out but can be thought of as a triangle becoming exceedingly obtuse and closing in on a straight line) which would correspond to an equilateral triangle.

I hope someone can finish the whole problem from here.