Difference between revisions of "1969 IMO Problems/Problem 6"
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==Problem== | ==Problem== | ||
− | Prove that for all real numbers <math>x_1, x_2, y_1, y_2, z_1, z_2</math>, with <math>x_1 > 0, x_2 > 0, x_1y_1 - z_1^2 > 0, x_2y_2 - z_2^2 > 0</math>, the inequality<cmath>\frac{8}{(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2} \leq \frac{1}{x_1y_1 - z_1^2} + \frac{1}{x_2y_2 - z_2^2}</cmath>is satisfied. Give necessary and sufficient conditions for equality. | + | Prove that for all real numbers <math>x_1, x_2, y_1, y_2, z_1, z_2</math>, with <math>x_1 > 0, x_2 > 0, |
+ | x_1y_1 - z_1^2 > 0, x_2y_2 - z_2^2 > 0</math>, the inequality | ||
+ | <cmath>\frac{8}{(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2} \leq \frac{1}{x_1y_1 - z_1^2} + \frac{1}{x_2y_2 - z_2^2}</cmath> | ||
+ | is satisfied. Give necessary and sufficient conditions for equality. | ||
==Solution== | ==Solution== | ||
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~Tomas Diaz. orders@tomasdiaz.com | ~Tomas Diaz. orders@tomasdiaz.com | ||
− | ==Solution | + | ==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: | This solution is actually more difficult but I added it here for fun to see the generalized case as follows: | ||
Line 89: | Line 92: | ||
~Tomas Diaz. orders@tomasdiaz.com | ~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 | ||
+ | <math>x_1y_1+x_2y_2-z_1z_1-z_2z_2 \le x_1y_2+x_2y_1-z_1z_2-z_1z_2</math> | ||
+ | is false. For a counterexample take <math>x_1 = y_1 = 2, x_2 = y_2 = 1, | ||
+ | z_1 = z_2 = 0.5</math>. The left hand side equals <math>5 - 0.25 - 0.25</math> | ||
+ | and the right hand side equals <math>4 - 0.25 - 0.25</math>. | ||
+ | |||
+ | 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 <math>y_1 > 0, y_2 > 0</math>. Also, we can assume <math>z_1 \ge 0, z_2 \ge 0</math>. | ||
+ | Indeed, if <math>z_1 < 0</math>, then <math>z_1 < -z_1</math>. It follows | ||
+ | <math>z_1 + z_2 < -z_1 + z_2</math>, so | ||
+ | <cmath>\frac{1}{(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2} < | ||
+ | \frac{1}{(x_1 + x_2)(y_1 + y_2) - (-z_1 + z_2)^2}</cmath> | ||
+ | |||
+ | So, if we proved the inequality for positive <math>z_1</math> it is also true | ||
+ | for negative <math>z_1</math>. | ||
+ | |||
+ | Now consider the function <math>F</math> of three variables | ||
+ | <math>F(x, y, z) = \frac{1}{xy - z^2}</math> defined on the domain | ||
+ | <math>x, y > 0, z \ge 0, z^2 < xy</math>. For simplicity, denote a | ||
+ | point <math>(x, y, z)</math> in the domain by <math>P</math>. | ||
+ | The inequality in the problem can be rewritten as | ||
+ | <cmath>F \left( \frac{P_1 + P_2}{2} \right) \le \frac{1}{2}[F(P_1) + F(P_2)]</cmath> | ||
+ | |||
+ | This follows immediately from the inequality expressing the | ||
+ | fact that <math>F(x, y, z)</math> is a convex function. (In fact, it | ||
+ | is equivalent to the convexity of <math>F</math>, but this is not needed | ||
+ | for our proof of the given inequality.) Therefore, it is enough | ||
+ | to prove that the function <math>F(x, y, z)</math> is convex. | ||
+ | |||
+ | We will prove that this function is convex. In fact, we will | ||
+ | prove that the function is \italic{strictly} convex. This will | ||
+ | imply that equality holds only when <math>P_1 = P_2</math>, in other words, | ||
+ | <math>x_1 = x_2, y_1 = y_2, z_1 = z_2</math>, which will give us the | ||
+ | necessary and sufficient conditions for equality. | ||
+ | |||
+ | [TO BE CONTINUED. SAVING, SO THAT I DON'T LOOSE WORK DONE SO FAR.] | ||
{{alternate solutions}} | {{alternate solutions}} | ||
== See Also == {{IMO box|year=1969|num-b=5|after=Last Question}} | == See Also == {{IMO box|year=1969|num-b=5|after=Last Question}} |
Revision as of 18:37, 30 July 2024
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 \italic{strictly} convex. This will
imply that equality holds only when , in other words,
, which will give us the
necessary and sufficient conditions for equality.
[TO BE CONTINUED. SAVING, SO THAT I DON'T LOOSE WORK DONE SO FAR.]
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 |