38 If $a, b, c>0, a \cos ^{2} \theta+b \sin ^{2} \theta<c$, then the size relationship between $\sqrt{a} \cos ^{2} \theta+\sqrt{b} \sin ^{2} \theta$ and $\sqrt{c}$ is
$\qquad$
$38<$. Because
$$
\begin{aligned}
& a \cos ^{2} \theta+b \sin ^{2} \theta-\left(\sqrt{a} \cos ^{2} \theta+\sqrt{b} \sin ^{2} \theta\right)^{2} \\
= & a \cos ^{2} \theta\left(1-\cos ^{2} \theta\right)-2 \sqrt{a b} \sin ^{2} \theta \cos ^{2} \theta+b \sin ^{2} \theta\left(1-\sin ^{2} \theta\right) \\
= & (\sin \theta \cos \theta)^{2}(\sqrt{a}-\sqrt{b})^{2} \geqslant 0,
\end{aligned}
$$

Therefore, $\left(\sqrt{a} \cos ^{2} \theta+\sqrt{b} \sin ^{2} \theta\right)^{2} \leqslant a \cos ^{2} \theta+b \sin ^{2} \theta<c$.