11. (20 points) Let $\alpha, \beta$ be any two interior angles of a triangle. If there exist $\alpha, \beta$ such that the inequality
$$
\cos \sqrt{\alpha}+\cos \sqrt{\beta}>a+\cos \sqrt{\alpha \beta}
$$

holds, find the range of the real number $a$.

保留了源文本的换行和格式。
11. From the problem, we know $\alpha, \beta>0$, and $0<\sqrt{\alpha}<1, \sqrt{\beta}>1$, then
$$
\begin{array}{l}
\sqrt{\alpha} \sqrt{\beta} \leqslant\left(\frac{\sqrt{\alpha}+\sqrt{\beta}}{2}\right)^{2} . \\
\text { Let } \frac{\sqrt{\alpha}+\sqrt{\beta}}{2}=t .
\end{array}
$$

From $(x+y)^{2} \leqslant 2\left(x^{2}+y^{2}\right)$
$\Rightarrow 0<2 \cos t$, that is
$$
f(t)=1+\cos t^{2}-2 \cos t>0\left(t \in\left(1, \sqrt{\frac{\pi}{2}}\right)\right) \text {. }
$$

Then $f^{\prime}(t)=2 t\left(\frac{\sin t}{t}-\sin t^{2}\right)$.
By $1<f\left(\sqrt{\frac{\pi}{2}}\right)>0$.
Thus, $\cos \sqrt{\alpha}+\cos \sqrt{\beta}-\cos \sqrt{\alpha \beta}<1$.
In summary, $\cos \sqrt{\alpha}+\cos \sqrt{\beta}-\cos \sqrt{\alpha \beta}<1$, and when $\sqrt{\alpha} \rightarrow 0, \sqrt{\beta} \rightarrow 0$,
$\cos \sqrt{\alpha}+\cos \sqrt{\beta}-\cos \sqrt{\alpha \beta} \rightarrow 1$.
Therefore, $a \in(-\infty, 1)$.