Jensen's inequality proof explanation I was reading a proof of Jensen's inequality on convex functions, and I need some help understanding it. The proof is as follows:
$$f(t_1x_1+...+t_nx_n) = f((1-t_n)(\frac{t_1}{1-t_n}x_1+...+\frac{t_{n-1}}{1-t_n}x_{n-1})+t_nx_n)$$
$$\le (1-t_n) f(\frac{t_1}{1-t_n}x_1+...+\frac{t_{n-1}}{1-t_n}x_{n-1})+t_nf(x_n)),(convexitivity)$$
$$\le (1-t_n) \{ \frac{t_1}{1-t_n}f(x_1)+...+\frac{t_{n-1}}{1-t_n}f(x_{n-1})\}+t_nf(x_n),(induction)$$
$$=t_1f(x_1)+...+t_nf(x_n)$$.
The proof is fairly simple but can someone please explain the inductive step for me.Thanks.

The proof write-up you read isn't friendly to those new to induction, partly because it's easy to mistake the induction for circularity, partly because the base step is a little non-standard (we have to check two values of $n$, not just one). So let me spell out the argument a little more.
Theorem: for any $n\in\Bbb N$, a convex $f$ satisfies$$\sum_{i=1}^nt_i=1,\,t_i\ge0\implies f\left(\sum_{i=1}^n t_i x_i\right)\le\sum_{i=1}^nt_i f(x_i).$$
Base step of inductive proof: if $n=1$ then $t_1=1$, so the required result is $f(x_1)\le f(x_1)$, which is true; if $n=2$ the required result is$$t_1+t_2=1,\,t_i\ge0\implies f(t_1x_1+t_2x_2)\le t_1f(x_1)+t_2f(x_2),$$which is true by the convexity of $f$.
Inductive step of inductive proof: we prove if the result holds for $n=k\ge 2$ then it also holds for $n=k+1$. If $t_{n+1}=1$ other $t_i$ are $0$, so the required inequality is $f(x_{n+1})\le f(x_{n+1})$, which is true. If $t_{n+1}\ne1$,$$f\left(\sum_{i=1}^{k+1}t_ix_i\right)=f\left((1-t_{k+1})\sum_{i=1}^k\frac{t_i}{1-t_{k+1}}x_i+t_{k+1}x_{k+1}\right).$$By convexity, we have the upper bound $$(1-t_{k+1})f\left(\sum_{i=1}^k\frac{t_i}{1-t_{k+1}}x_i\right)+t_{k+1}f(x_{k+1}).$$By the inductive hypothesis (i.e. the $n=k$ case of the result), we get an upper bound on the first term, so$$f\left(\sum_{i=1}^{k+1}t_ix_i\right)\le(1-t_{k+1})\sum_{i=1}^k\frac{t_i}{1-t_{k+1}}f(x_i)+t_{k+1}f(x_{k+1})=\sum_{i=1}^{k+1}t_if(x_i).$$Our inductive step is now complete.
