<s>[INST] Verify that $\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2})\le\frac{1}{8}$ in a general triangle $\triangle ABC$ So, this problem is inspired by a contest preparation problem I saw back in Japan, and it is as follows:

In a general triangle $\triangle ABC$, show that $\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2})\le\frac{1}{8}$

Now, while I still haven't figured out a geometric interpretation of this inequality, here is my attempt to prove this:
Recall that:
$$\sin^2(\frac{A}{2})=\frac{(1-\cos(A))}{2}$$
$$\sin^2(\frac{A}{2})=\frac{1}{2}(1-\frac{b^2+c^2-a^2}{2bc})$$
$$\sin^2(\frac{A}{2})=\frac{1}{2}(\frac{a^2-b^2-c^2+2bc}{2bc})$$
$$\sin^2(\frac{A}{2})=\frac{a^2-(b-c)^2}{4bc}$$
Now, obviously $\frac{a^2-(b-c)^2}{4bc} \le \frac{a^2}{4bc}$, therefore:
$$\sin^2(\frac{A}{2}) \le \frac{a^2}{4bc}$$
$$\sin(\frac{A}{2}) \le \frac{a}{2\sqrt{bc}}$$
This can be done for $\sin(\frac{A}{2}), \sin(\frac{B}{2})$ and $\sin(\frac{C}{2})$.
$$\sin(\frac{A}{2}) \le \frac{a}{2\sqrt{bc}}$$
$$\sin(\frac{B}{2}) \le \frac{b}{2\sqrt{ac}}$$
$$\sin(\frac{C}{2}) \le \frac{c}{2\sqrt{ab}}$$
Therefore:
$$\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2}) \le (\frac{a}{2\sqrt{bc}})(\frac{b}{2\sqrt{ac}})(\frac{c}{2\sqrt{ab}})$$
$$\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2}) \le \frac{abc}{8abc}$$
$$\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2}) \le \frac{1}{8}$$
However I'm not sure if this is correct or if it is, I don't think this brute force approach is good. Are there any better options to prove this inequality? Please share your answers!
 [/INST] By your work
$$\prod_{cyc}\sin\frac{\alpha}{2}=\prod_{cyc}\sqrt{\frac{(a-b+c)(a+c-b)}{4bc}}=\frac{\prod\limits_{cyc}(a+b-c)}{8abc}\leq\frac{1}{8}$$
because $$abc-\prod\limits_{cyc}(a+b-c)=\sum_{cyc}(a^3-a^2b-a^2c+abc)=$$
$$=\sum_{cyc}(a^3-abc-b^2c-a^2c+2abc)=\sum_{cyc}(a-b)^2\left(\frac{a+b+c}{2}-c\right)=$$
$$=\frac{1}{2}\sum_{cyc}(a-b)^2(a+b-c)\geq0.$$
By the way, last inequality is true for any non-negatives $a$, $b$ and $c$, but it's another problem.
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