The angles of a triangle are $\alpha, \beta, \gamma$. What is the maximum possible value of

$$
\sin \alpha \sin \beta \cos \gamma+\sin ^{2} \gamma
$$
I. solution. If $\alpha=\beta=\gamma=60^{\circ}$, then the value of the expression is

$$
\sin 60^{\circ} \sin 60^{\circ} \cos 60^{\circ}+\sin ^{2} 60^{\circ}=\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}=\frac{9}{8}
$$

We will show that in any other case, the value of the expression is smaller. Since $\alpha+\beta=180^{\circ}-\gamma$, it follows that $\cos (\alpha+\beta)=$ - $\cos \gamma$, and from the addition theorems, we get that

$$
\sin \alpha \sin \beta=\frac{\cos (\alpha-\beta)-\cos (\alpha+\beta)}{2}
$$

By transforming the expression using the above relationships, we obtain

$$
\begin{aligned}
K & =\sin \alpha \sin \beta \cos \gamma+\sin ^{2} \gamma=\frac{1}{2}(\cos (\alpha-\beta)-\cos (\alpha+\beta)) \cdot \cos \gamma+1-\cos ^{2} \gamma= \\
& =\frac{1}{2}(\cos (\alpha-\beta)+\cos \gamma) \cdot \cos \gamma+1-\cos ^{2} \gamma= \\
& =\frac{1}{2} \cos (\alpha-\beta) \cos \gamma+1-\frac{1}{2} \cos ^{2} \gamma= \\
& =1-\frac{1}{2}\left(\cos ^{2} \gamma-\cos (\alpha-\beta) \cos \gamma\right)= \\
& =1-\frac{1}{2}\left[\left(\cos \gamma-\frac{\cos (\alpha-\beta)}{2}\right)^{2}-\frac{\cos ^{2}(\alpha-\beta)}{4}\right]= \\
& =1+\frac{1}{8} \cos ^{2}(\alpha-\beta)-\frac{1}{2}\left(\cos \gamma-\frac{\cos (\alpha-\beta)}{2}\right)^{2}
\end{aligned}
$$

Since the square of any real number is non-negative, and for any real number $x$, $\cos ^{2} x \leq 1$, it follows that

$$
K \leq 1+\frac{1}{8} \cos ^{2}(\alpha-\beta) \leq \frac{9}{8}
$$

It is also clear that equality holds precisely when $\alpha=\beta$ and $\cos \gamma=\frac{\cos 0^{\circ}}{2}=\frac{1}{2}$, i.e., when the triangle is equilateral. This proves the statement.

II. solution. Similarly to the first solution, we start from the relationship $\alpha+\beta=180^{\circ}-\gamma$. This implies that

$$
\sin \gamma=\sin (\alpha+\beta) \quad \text { and } \quad \cos \gamma=-\cos (\alpha+\beta)
$$

Using the addition formulas, we get:

$$
\begin{aligned}
K & =\sin \alpha \sin \beta \cos \gamma+\sin ^{2} \gamma= \\
& =-\sin \alpha \sin \beta(\cos \alpha \cos \beta-\sin \alpha \sin \beta)+(\sin \alpha \cos \beta+\cos \alpha \sin \beta)^{2}= \\
& =\sin ^{2} \alpha \sin ^{2} \beta+\sin ^{2} \alpha \cos ^{2} \beta+\cos ^{2} \alpha \sin ^{2} \beta+\sin \alpha \sin \beta \cos \alpha \cos \beta
\end{aligned}
$$

Since for any angle $x$, $\sin ^{2} x+\cos ^{2} x=1$, it follows that

$$
\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)\left(\sin ^{2} \beta+\cos ^{2} \beta\right)=1
$$

Thus,

$$
\sin ^{2} \alpha \sin ^{2} \beta+\sin ^{2} \alpha \cos ^{2} \beta+\cos ^{2} \alpha \sin ^{2} \beta=1-\cos ^{2} \alpha \cos ^{2} \beta
$$

Therefore,

$$
\begin{aligned}
K & =1-\cos ^{2} \alpha \cos ^{2} \beta+\sin \alpha \sin \beta \cos \alpha \cos \beta= \\
& =1+\cos \alpha \cos \beta(\sin \alpha \sin \beta-\cos \alpha \cos \beta)=1+\cos \alpha \cos \beta \cos \gamma
\end{aligned}
$$

In a triangle, the product of the cosines of the angles is positive if and only if the triangle is acute. Therefore, if we are looking for the maximum of $K$, we can assume that the triangle is acute. Since the $\cos x$ function is positive and concave from below on the interval $\left(0, \frac{\pi}{2}\right)$, by the arithmetic and geometric mean inequality and Jensen's inequality, we have

$$
K \leq 1+\left(\frac{\cos \alpha+\cos \beta+\cos \gamma}{3}\right)^{3} \leq 1+\cos \left(\frac{\alpha+\beta+\gamma}{3}\right)^{3}=1+\cos ^{3} 60^{\circ}=\frac{9}{8}
$$

Thus, the maximum value of $K$ is $\frac{9}{8}$, and it is attained precisely when the triangle is equilateral.