proving f(n) is strictly less than 2.4 when $n\geq 1$ $f(1)= 2, f(n+1) =  \sqrt{3+f(n)}$.
Prove $f(n) < 2.4$ for all $n ≥ 1$.
Would this be a proof by induction? If so, could somebody start me off?

Hint: $\alpha=\frac{1}{2}\left(1+\sqrt{13}\right)$ is the only solution of $x=\sqrt{3+x}$. If you prove that
$$ \forall x\geq 2,\qquad  \left|\sqrt{3+x}-\alpha\right|\leq \left|x-\alpha\right| \tag{1} $$
it follows that for every $n\geq 3$
$$ f(n)\in [2,\alpha]\tag{2} $$
since $\sqrt{3+x}\geq x$ on $[2,\alpha]$.
You may notice that $(1)$ can be written as
$$\left|\frac{\sqrt{3+x}-\sqrt{3+\alpha}}{x-\alpha}\right|\leq 1 \tag{3} $$
for any $x\neq\alpha$, and that $(1)$ follows from the Lipshitz-continuity of $\sqrt{3+x}$ over $[2,+\infty)$.
Actually $(1)$ can be improved by much:
$$\forall x\geq 2,\qquad \left|\sqrt{3+x}-\alpha\right|\leq\color{red}{\frac{1}{2\sqrt{5}}}\left|x-\alpha\right|.\tag{1improved}$$
