What Is The Smallest Real Number?

The answer is $\frac {5} {4}$ .

Let us prove that

$f = \sin^2 \alpha + \sin^2 \beta - \cos \gamma < \frac{5}{4} \hspace{30pt} \text{(1)}$

Indeed, $f=2- \cos^2 \alpha - \cos^2 \beta - \cos \gamma = 1 - \frac{1}{2} (\cos 2\alpha + \cos 2\beta) + \cos(\alpha + \beta) = 1 - (\cos (\alpha + \beta)(\cos(\alpha-\beta) - 1)$

and the inequality $\text{(1)}$ is equivalent to

$(\cos(\alpha + \beta)(1-\cos(\alpha-\beta))<\frac{1}{4}\hspace{30pt}\text{(2)}$

$\text{(2)}$ follows from the inequality $ab \le \frac{(a+b)^2}{4}$ .

Indeed, $(\cos(\alpha+\beta)(1-\cos(\alpha-\beta))\le \frac{1}{4}(\cos(\alpha+\beta)+1-\cos(\alpha-\beta))^2=\frac{1}{4}(1-2 \sin \alpha \sin \beta)^2 < \frac{1}{4}$ , since $0< \sin \alpha \sin \beta < 1$ ( $\alpha, \beta$ are triangle angles).

$\text{(1)}$ is proved. Now note that if $\gamma=\frac{2\pi}{3}$ and $\alpha$ approaches to $\frac{\pi}{3}$ , then {f} approaches to $\frac{5}{4}$ .

$\frac{5}{4} = \boxed{1.25}$