10. Given positive real numbers $a, b$ satisfy: for real number $x$, when $1 \leqslant x \leqslant 4$, we have $a x+b-3 \leqslant 0$. Then the minimum value of $\frac{1}{a}-b$ is $\qquad$
10. 1 .

Let $f(x)=a x+b-3$. Then $f(x)$ is a linear function. From the conditions, we have
$$
\left\{\begin{array}{l}
a+b-3 \leqslant 0, \\
4 a+b-3 \leqslant 0
\end{array} \Rightarrow a \leqslant \frac{1}{4}(3-b)\right. \text {. }
$$

Also, since $a>0$, we know $0<b<3$.
Thus, $\frac{1}{a}-b \geqslant \frac{4}{3-b}-b=\frac{4}{3-b}+(3-b)-3$
$$
\geqslant 2 \sqrt{\frac{4}{3-b}(3-b)}-3=1 \text {. }
$$

When $b=1, a=\frac{1}{2}$, the equality holds.
Therefore, the minimum value of $\frac{1}{a}-b$ is 1.