113 Let $a, b$ be real numbers satisfying $a>b \geqslant 0$, then the maximum value of the function $y=(a-b) \sqrt{1-x^{2}}+a x$ on $-1 \leqslant$ $x \leqslant 1$ is $\qquad$ .
$113 \sqrt{(a-b)^{2}+a^{2}}$. Since $-1 \leqslant x \leqslant 1$, we can set $x=\cos \theta, (0 \leqslant \theta \leqslant \pi)$. Therefore, we have
$$
y=(a-b) \sin \theta + a \cos \theta = \sqrt{(a-b)^{2}+a^{2}} \sin (\theta + \alpha).
$$

where
$$
\sin \alpha = \frac{a}{\sqrt{(a^{2}-b^{2})+a^{2}}}, \left(0 < \alpha < \frac{\pi}{2}\right).
$$

Thus,
$$
y_{\text{max}} = \sqrt{(a-b)^{2}+a^{2}} \sin 90^{\circ} = \sqrt{(a-b)^{2}+a^{2}}.
$$

Or by the Cauchy inequality, $y^{2} \leqslant \left((a-b)^{2}+a^{2}\right)\left(1-x^{2}+x^{2}\right) = (a-b)^{2}+a^{2}$.