Problem 1330

13.13. Let $\alpha_{1}=\angle O A C, \quad \alpha_{2}=\angle O A B, \quad \beta_{1}=\angle O B A$, $\beta_{2}=\angle O B C, \quad \gamma_{1}=\angle O C B, \quad \gamma_{2}=\angle O C A, \quad \alpha=\alpha_{1}+\alpha_{2}, \quad \beta=\beta_{1}+\beta_{2}$,

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Fig. 84

$\gamma=\gamma_{1}+\gamma_{2}$ (Fig. 84). Then the required inequality follows from the following chain of relations:

$p=\frac{a+b+c}{2}=\frac{1}{2}\left(O C \cos \gamma_{1}+O C \cos \gamma_{2}+O B \cos \beta_{1}+O B \cos \rho_{2}+\right.$

$$\begin{aligned} & \left.+O A \cos \alpha_{1}+O A \cos \alpha_{2}\right)=O C \cos \frac{\gamma}{2} \cos \frac{\gamma_{1}-\gamma_{2}}{2}+ \\ & +O B \cos \frac{\beta}{2} \cos \frac{\beta_{1}-\beta_{2}}{2}+O A \cos \frac{\alpha}{2} \cos \frac{\alpha_{1}-\alpha_{2}}{2} \leqslant \\ & \quad \leq O A \cos \frac{\alpha}{2}+O B \cos \frac{\beta}{2}+O C \cos \frac{\gamma}{2} \end{aligned}$$

Equality is achieved when

$$\alpha_{1}=\alpha_{2}, \quad \beta_{1}=\beta_{2}, \quad \gamma_{1}=\gamma_{2},$$

i.e., when point $O$ is the intersection of the angle bisectors of triangle $A B C$.

13:14. Note that when $\alpha=\beta=\gamma=60^{\circ}$, the required inequality becomes an equality. Suppose at least two angles of the triangle, say $\alpha$ and $\beta$, are not equal, then

$\cos \alpha+\cos \beta+\cos \gamma=\cos \alpha+\cos \beta-\cos (\alpha+\beta)=$

$=2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}-\cos (\alpha+\beta)<2 \cos \frac{\alpha+\beta}{2}-2 \cos ^{2} \frac{\alpha+\beta}{2}+1=$ $=-\frac{1}{2}\left(2 \cos \frac{\alpha+\beta}{2}-1\right)^{2}+\frac{3}{2} \leqslant \frac{3}{2}$, i.e., the required inequality is also satisfied but is strict. Thus, equality holds only for an equilateral triangle.