Derive Hoeffding's inequality from Hoeffding's lemma

Assuming Hoeffding's lemma — a centred variable supported on $[a,b]$ satisfies $\E e^{\lambda X}\le e^{\lambda^2(b-a)^2/8}$ — derive the full Hoeffding inequality

$$\P\Big(\sum_{i=1}^n(X_i-\E X_i) \ge t\Big) \le \exp\left(-\frac{2t^2}{\sum_i(b_i-a_i)^2}\right),$$

carrying out the optimisation over $\lambda$ explicitly. Then invert it to state the bound in "with probability $1-\delta$" form for i.i.d. variables in $[0,1]$.