Integrate the inequality from $t = 0$ to $t=1$ gives
$$L(x, a+b) - L(x, a) \le c|b| L(x,a) \Rightarrow L(x, a+b) \le (c|b| +1) L(x,a).$$
Now the result follows the inequality $e^x \ge 1+x$ for $x\ge 0$.
Integrate the inequality from $t = 0$ to $t=1$ gives
$$L(x, a+b) - L(x, a) \le c|b| L(x,a) \Rightarrow L(x, a+b) \le (c|b| +1) L(x,a).$$
Now the result follows the inequality $e^x \ge 1+x$ for $x\ge 0$.