1969 IMO Problems/Problem 6
Contents
[hide]Problem
Prove that for all real numbers , with , the inequality is satisfied. Give necessary and sufficient conditions for equality.
Solution
Let and
From AM-GM:
with equality at
[Equation 1]
since and , and using the Rearrangement inequality
then
[Equation 2]
Therefore, we can can use [Equation 2] into [Equation 1] to get:
Then, from the values of and we get:
With equality at and
~Tomas Diaz. orders@tomasdiaz.com
Generalization and Idea for a Solution
This solution is actually more difficult but I added it here for fun to see the generalized case as follows:
Prove that for all real numbers , for with
and the inequality
is satisfied.
Let and
From AM-GM:
with equality at
[Equation 3]
Here's the difficult part where I'm skipping steps:
we prove that
and replace in [Equation 3] to get:
and replace the values of and to get:
with equality at for all
Then set and substitute in the generalized inequality to get:
with equality at
~Tomas Diaz. orders@tomasdiaz.com
Remarks (added by pf02, July 2024)
1. The solution given above is incorrect. The error is in the incorrect usage of the Rearrangement inequality. The conclusion is false. For a counterexample take . The left hand side equals and the right hand side equals .
2. The generalization is reasonable but the idea for a solution is unacceptably vague (at one crucial step, the author says "Here's the difficult part where I'm skipping steps"). I don't believe this can be developed into a real proof, since it just follows the idea of the Solution above, which is incorrect.
3. I will give a solution below, which uses calculus. I believe an "elementary" solution (i.e. a solution based on elementary algebra and geometry) is possible, but quite difficult.
Solution
First, remark that given the conditions of the problem, it follows that . Also, we can assume . Indeed, if , then . It follows , so
So, if we proved the inequality for positive it is also true for negative .
Now consider the function of three variables defined on the domain . For simplicity, denote a point in the domain by . The inequality in the problem can be rewritten as
This follows immediately from the inequality expressing the fact that is a convex function. (In fact, it is equivalent to the convexity of , but this is not needed for our proof of the given inequality.) Therefore, it is enough to prove that the function is convex.
We will prove that this function is convex. In fact, we will prove that the function is strictly convex. This will imply that equality holds only when , in other words, , which will give us the necessary and sufficient conditions for equality.
We use the theorem which says that a twice differentiable function defined on a convex set is convex if and only if its Hessian is positive definite. Furthermore, if the Hessian is strictly positive definite, then the function is strictly convex.
The Hessian of the function is the 3x3 matrix formed by the second derivatives of :
is positive definite if for any 3x1 vector we have . ( is the transpose of , a matrix which is 1x3.) is strictly positive definite if unless .
Explicitly, we need to show that if are three numbers, then
and that the expression is only when .
Before embarking on the computations proving the inequality expressing the positive definiteness of the Hessian of , let us show that the domain of is convex. Let and be two points in the domain of . This means that , , and . We need to verify that the same is true for when . The only thing which is not obvious is . To see this, work out the multiplications, use what we already know about and , simplify by and we are left with showing that . This can be seen as follows: .
Now work towards proving the inequality expressing that the Hessian of is positive definite. Compute all the second derivatives of ; notice that they all have at the denominator, so we can simplify with this factor. (The computation is made somewhat shorter by taking advantage of the fact that the order of taking the derivatives makes no difference in mixed derivatives (for example ).)
We get (as the thing we have to prove) that
.
If the expression equals if and only if as well. So, assume that at least one of is . We can assume . We will show that in this case, the expression is .
Simplify by , and rearrange the terms so we can think of the expression on the left as a polynomial in :
.
Notice that this is a polynomial (in ) of degree of the type , whose leading coefficient is . To show that its values are always we need to show that its discriminant is , in other words, .
Therefore, we need to show that
After factoring out , and working out all multiplications, this becomes
After some rearranging of terms, grouping, and factoring, this becomes
Since , we need to show
Now, view the expression on the left hand side as a polynomial of degree in , whose leading coefficient is . To conclude that its values are we need to show that its discriminant is . That is, we need to show that
Work out the multiplications, and regroup terms to get
.
This is clearly true, which concludes the proof.
[Solution by pf02, July 2024]
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1969 IMO (Problems) • Resources | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Last Question |
All IMO Problems and Solutions |