It is possible to prove this inequality by first applying Law of Cosines to make it an inequality with respect to $a$, $b$, and $c$. Alternatively, it can also be proved by working on its corresponding equality (see # 3879) $$\cos A +\cos B + \cos C = 1+4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$$ By %%HREF%%3880%%, we have $$\sin\frac{A}{2}=\sqrt{\frac{(p-b)(p-c)}{bc}}$$ Similarly, we have $$\sin\frac{B}{2}=\sqrt{\frac{(p-c)(p-a)}{ca}}$$ and $$\sin\frac{C}{2}=\sqrt{\frac{(p-a)(p-b)}{ab}}$$ Multiplying these three relations together yields $$\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}=\frac{(p-a)(p-b)(p-c)}{abc}$$ By Heron's formula, we find $$(p-a)(p-b)(p-c)=\frac{S^2_{\triangle{ABC}}}{pabc}$$ Therefore, the to-be-proved claim is equivalent to $$r=4R\cdot\frac{S^2_{\triangle{ABC}}}{p\cdot abc}\quad\Leftrightarrow\quad rp=S_{\triangle{ABC}}\frac{4R\cdot S_{\triangle{ABC}}}{abc}$$ Applying Heron's formula again find the left side of the above relation equals to $S_{\triangle{ABC}}$. And we know $4R\cdot S_{\triangle{ABC}}=abc$ (see # 3881). Hence, the above relation does hold.