I am attempting to prove that if $f$ is a convex function, then $\forall x_{1},...,x_{m}\in\mathbb{R}^n, \forall t_{1},...,t_{m}\geq0:t_{1}+...+t_{m}=1$, we have $f(t_{1}x_{1}+...+t_{m}x_{m})\leq t_{1}f(x_{1})+...+t_{m}f(x_{m})$.
My initial attempt was to show this by induction, where it is clear that for $m=1$, this is trivial. $f(t_{1}x_{1})\leq t_{1}f(x_{1})$, since $f$ is convex. As well as for $m=2$, $f(t_{1}x_{1}+t_{2}x_{2})\leq t_{1}f(x_{1})+t_{2}f(x_{2})$, because $f$ is convex. My induction hypothesis is that this holds for $k\leq m$.
$f(t_{1}x_{1}+...+t_{k}x_{k})\leq t_{1}f(x_{1})+...+t_{k}f(x_{k})$
However, I do not see how I can use my hypothesis for proving this holds for $m=k+1$. Is there some other definition of $f$ being convex that I should use. I do not think we can assume $f$ is linear.
Fix $x_1,\ldots,x_{k+1}$ and $t_1,\ldots,t_{k+1}$ such that $\sum_{i=1}^{k+1}t_i=1$.
Write $(1-t_{k+1})y=t_1x_1+\cdots+t_kx_k$.
By convexity we have that $$ f((1-t_{k+1})y+t_{k+1}x_{k+1}) \le (1-t_{k+1})f(y) +t_{k+1}f(x_{k+1}) $$
By the induction hypothesis, $$ f(y) = f\left( \frac{t_1}{1-t_{k+1}}x_1 +\cdots +\frac{t_k}{1-t_{k=1}}x_k\right) \le \frac{t_1}{1-t_{k+1}} f\left( x_1\right) +\cdots +\frac{t_k}{1-t_{k+1}}f\left(x_k\right) $$
which implies that $$ f(t_1x_1 + \cdots + t_{k+1}x_{k+1}) \le t_1f(x_1)+\cdots+t_{k+1}f(x_{k+1}) $$