Let $f$ be defined on an open interval $I := (a,b)$.
(a) Let $x$ and $y$ be real numbers such that $x<y$. Show that if $z \in [x,y]$, then there is some $t \in [0,1]$ such that $z=tx+(1-t)y$.
(b) Assume that $f$ satisfies the following conditions: $$\forall x,y \in (a,b), \forall t \in [0,1], f(tx+(1-t)y) \leq tf(x)+(1-t)f(y) \tag{1}$$ (in other words $f$ is convex). Show that for any positive integer m, whenever $x_1,...,x_m$ are in $I$ and $t_1,...,t_m$ are non-negative numbers with $t_1+...+t_m=1$, then, $f(t_1x_1+...+t_mx_m)\leq t_1f(x_1)+...+t_mf(x_m)$
(c) Suppose that $f'$ is increasing on $I$. Show that $f$ satisfies (1).
(d) Let $g(x) := ln x$ for $x>0$. Put $f:=-g$. Show that $f'$ is increasing on $(0,\infty)$. Hence deduce that if $m \in \mathbb N, x_1,...,x_m$ are positive and $t_1,...,t_m$ are non-negative numbers with $\sum_{k=1}^m t_k=1$, then $$x_1^{t_1}...x_m^{t_m}\leq t_1x_1+...+t_mx_m. \tag{2}$$
Put $t_k:=1/m$ for all $k=1,...,m$ in (2). What do you obtain?
What I think should be done:
I am not sure what I should be doing for parts (a) and (b). Could anyone help me out please? I think we should apply the MVT here but I am not sure how exactly.
(c) So, I understand that if $x<y$ and $z=tx+(1-t)y$, then we can apply the MVT on $[x,z]$ and $[z,y]$ but I am not sure how to do that to show that $f'$ is increasing.
(d) So, here, $f(x)=-lnx$. So, $f'(x)=-1/x$. I think we should use the following corollary: Suppose that $f$ is continuous on [a,b] and $f'(x)>0$ for all $x \in (a,b)$, then f is strictly increasing on $[a,b]$. That is, if $x_1,x_2$ are in $[a,b]$ and $x_1<x_2$, then $f(x_1)<f(x_2)$. How can I use this here? Also, From there, how can I deduce the rest of the question?
Thanks