Inequalities for convex functions

A convex function $f(x)$ on an interval $[a,b]$ is maximized at $x=a$ or $x=b$ (or maybe both).

Example (USAMO 1980/5):
Prove that for $a,b,c \in [0,1]$,

$$\frac{a}{b+c+1} + \frac{b}{c+a+1} +\frac{c}{a+b+1} + (1-a)(1-b)(1-c) \ge 1.$$

Solution:
Let $F(a,b,c)$ denote the left hand side. If we fix $b$ and $c$ in $[0,1]$, the resulting function of $a$ is convex on $[0,1]$, because it is a sum of functions of the type $f(a)=r/(s+a)$ and linear functions. Therefore it is maximized when $a=0$ or $a=1$; i.e., we can increase $F(a,b,c)$ by replacing $a$ by 0 or $1$. Similarly one can increase $F(a,b,c)$ by replacing each of $b$ and $c$ by 0 or $1$. Hence the maximum value of $F(a,b,c)$ will occur at one of the eight vertices of the cube $0 \le a,b,c \le 1$. But $F(a,b,c)=1$ at these eight points, so $F(a,b,c) \le 1$ whenever $0 \le a,b,c \le 1$.

Jensen's Inequality:
Let $f$ be a convex function on an interval $I$. If $x_1,\dots,x_n \in I$, then

$$\frac{f(x_1)+\dots+f(x_n)}{n} \ge f\left(\frac{x_1 x_2 \dots x_n}{n}\right).$$

If moreover $f$ is strictly convex, then equality holds if and only if $x_1=x_2=\dots=x_n$.

Hardy-Littlewood-Polyà majorization inequality:
Let $f$ be a convex function on an interval $I$, and suppose $a_1,\dots,a_n,b_1,\dots,b_n \in I$. Suppose that the sequence $a_1,\dots,a_n$ majorizes $b_1,\dots,b_n$: this means that the following hold:

$$a_1 \ge \dots \ge a_n$$

$$b_1 \ge \dots \ge b_n$$

$$a_1 \ge b_1$$

$$a_1+a_2 \ge b_1+b_2$$

$$\vdots$$

$$a_1+a_2+\dots+a_{n-1} \ge b_1+b_2+\dots+b_{n-1}$$

$$a_1+a_2+\dots+a_{n-1}+a_n = b_1+b_2+\dots+b_{n-1}+b_n.$$

(Note the equality in the final equation.) Then

$$f(a_1)+\dots+f(a_n) \ge f(b_1)+\dots+f(b_n).$$

If $f$ is strictly convex on $I$, then equality holds if and only if $a_i=b_i$ for all $i$.