1.1 Convex Functions and Jenson’s Inequality
A function f is called a convex function if the line segment between any two points on the graph of the function lies above or on the graph. The following curve illustrates this oncept of a convex function and concave function:
Fig 1.1
For a convex function f, we have a famous inequality called Jenson’s Inequality:
For x_1,x_2,…,x_n \in \mathbb{R}, and a_1,a_2,…a_n\ge 0 with a_1+a_2…a_n=1, we have
f(a_1x_1+a_2x_2+…a_nx_) \le a_1f(x_1)+a_2f(x_2)+…a_nf(x_n).
The proof of Jensen’s inequality is very simple. Since the graph of every convex function lies above its tangent line at every point, we can compare the function f by l whose graph is tangent to the graph of a_1x_1+a_2x_2+…+a_nx_n. Then the left hand side of the inequality is the same for f and l, while the right hand side is smaller for l. But the equality case holds for all linear functions!
Jenson’s inequality has a lot of applications including the AM-GM inequality being one of them! We can derive AM-GM inequality from Jenson’s Inequality by considering the function f(x)=-\ln x.
1 important result of convexity should be observed over here. For convex f and reals a_i the following inequality holds true:
\frac{f(a_1)+f(a_2)+…f(a_n)}{n} \ge f\left(\frac{a_1+a_2+..a_n}{n}\right)
1.2 Triangle Inequalitites
We often deal with inequalities regarding sides of a triangle. We’ll keep handy the following facts about triangle inequalitites, with a,b,c as the sides of the triangle:
- a+b>c, b+c>a, a+c>b. Also, if no other conditions are specified it makes sense to assume a\le b\le c.
- a>|b-c|, b>|a-c|, c>|a-b|
- (a+b-c)(b+c-a)(c+a-b)>0
- a=y+z, b=z+x, c=x+y with x,y,z \in \mathbb{R}^+
2.1 Power Mean Inequality
Suppose that we have positive reals a_1,a_2,..a_n and reals k_1,k_2 with k \ge k_2, we have the following inequality:
The proof of this follows from the fact that f(x)=x^{\frac{k_2}{k_1}} is concave for x>0,and so we can apply Jenson’s Inequality to get
2.2 Cauchy-Schwarz Inequality
This is one of the most ueful, but most underrated inequality of all times:
For reals a_i,b_i
(a_1b_1+a_2b_2+…+a_nb_n)^2 \le (a_1^2+a_2^2+….a_n^2)(b_1^2+…+b_n^2).
Putting b_i=1 for all i we have
\left(\sum\limits_{i=1}^na_i\right)^2 \le n\sum\limits_{i=1}^na_i^2
How do you prove the last CS inequality?