Show that $f({x_1+x_2+\cdots x_n\over n})\le \frac 1n [f(x_1)+f(x_2)+\cdots f(x_n)]$ If $f''(x)\ge 0$ on $[a,b]$ prove that $f({x_1+x_2+\cdots x_n\over n})\le \frac 1n[f(x_1)+f(x_2)+\cdots f(x_n)]$
for any n points $x_1,x_2,...,x_n$ in $[a,b]$.
Applying MVT on $[x_1,{x_1+x_2\over 2}]$ & $[{x_1+x_2\over 2},x_2]$ and using the fact $f'$ is increasing on $[a,b]$, I could prove the proposition for $n=2$. But I couldn't use induction after that.

This is a special caes of Jensen's inequality. Standard proofs are given on this wiki page.
It is also possible to prove this special case by a slightly unusual induction.
We let $P(n)$ denote the proposition for $n$. You have already shown that $P(2)$ is true.
Now the proof goes like this:

*

*We show that, for $n \geq 2$, if $P(n)$ is true, then $P(2n)$ is true.

*We show that, for $n \geq 2$, if $P(n)$ is true, then $P(n - 1)$ is true.

These two statements combined together will show that $P(n)$ is true for all $n$.

For 1, we have $$f(\frac{x_1 + \dots x_{2n}}{2n}) \leq \frac 1 2[f(\frac{x_1 + \dots + x_n}n) + f(\frac{x_{n + 1} + \dots + x_{2n}}n)] \leq \frac 1 {2n}[f(x_1) + \dots + f(x_{2n})]$$ where the first inequality uses $P(2)$ and the second inequality uses $P(n)$.

For 2, we put $x_n = \frac {x_1 + \dots + x_{n - 1}}{n - 1}$ in $P(n)$ and get $P(n - 1)$.
