Jensen's inequality is an inequality involving convexity of a function. We first make the following definitions:
A function is convex on an interval if the segment between any two points taken on its graph in lies above the graph. An example of a convex function is .
A function is concave on an interval if the segment between any two points on the graph in lies below the graph. An example of a concave function is .
Now, Jensen's inequality is the following:
Let a real-valued function be convex on the interval . Let and . Then we have
If is concave, the direction of inequality is flipped.
In particular if we take weights , we get the inequality
This proof is a little long. As it doesn't help much in the problem-solving aspect, you may skip it.
Note that we need only to prove the statement for convex functions. This is due to the following result:
If is a concave function, then is a convex function.
We first prove the following statement :
where and are as follows:
: and the following holds :
: and the following holds :
Simply put where
This means that we've reduced the problem to just showing that is true. To do this, we use induction.
The base case is trivially true.
The case is the definition of a convex function (the proof of this is left to the interested reader).
For , we shall assume that if the assertion trivially holds, and if we appeal to the induction hypothesis).
Now, we proceed assuming that the cases and are true :
So, the proof is complete via induction.
The last form is the form that is most frequently encountered and used in various Olympiad level problems.
How do we check if a function is convex or concave? We can't always plot the graph and check. The best (and often quicker) way is using calculus. A function is convex on the interval if and only if for all and concave if for all . The precise statement is given below:
Let be a twice-differentiable function.
- is convex on if and only if
- is concave on if and only if
The proof of this can be found here.
For example, for we have so for all . So the function is convex everywhere on the graph. Similarly for we have for all . So the function is concave everywhere on the graph.
Since we've been talking so much about , let's apply Jensen's on it. We've already shown that it's convex everywhere. We choose reals . Applying Jensen's we get
AM-GM inequality (arithmetic mean-geometric mean inequality) is one of the special cases of Jensen's inequality:
Consider the function
Observe that . Then turns out to be a concave function. Also notice that we can conclude is concave using the graph of Thus, by Jensen's inequality we have
By the property of logarithms, and . Therefore, we can simplify the terms on the RHS as
Taking antilogarithm, we get our desired result.
Prove that for all
Notice that on the left side denoted we have a sum of same terms with some different arguments, so we should think of applying Jensen's with the common form as a function. Define the function We can find the second derivative, but that's not needed if we observe that for , we have . So the graph should look like that of , V-shaped with a little curve at the bottom and thus convex.
Now we are ready to apply Jensen's with reals . We get
Hence, completing the proof.