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Difference between revisions of "1969 IMO Problems/Problem 6"
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We will prove that this function is convex.  In fact, we will
 
We will prove that this function is convex.  In fact, we will
−
prove that the function is \italic{strictly} convex.  This will
+
prove that the function is strictly convex.  This will
 
imply that equality holds only when <math>P_1 = P_2</math>, in other words,
 
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
 
<math>x_1 = x_2, y_1 = y_2, z_1 = z_2</math>, which will give us the

Revision as of 17:43, 30 July 2024

Problem

Prove that for all real numbers $x_1, x_2, y_1, y_2, z_1, z_2$, with $x_1 > 0, x_2 > 0, x_1y_1 - z_1^2 > 0, x_2y_2 - z_2^2 > 0$, the inequality \[\frac{8}{(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2} \leq \frac{1}{x_1y_1 - z_1^2} + \frac{1}{x_2y_2 - z_2^2}\] is satisfied. Give necessary and sufficient conditions for equality.

Solution

Let $A=x_1y_1 - z_1^2>0$ and $B=x_2y_2 - z_2^2>0$

From AM-GM:

$\sqrt{AB} \le \frac{A+B}{2}$ with equality at $A=B$

$4AB \le (A+B)^2$

$\frac{4}{A+B} \le \frac{A+B}{AB}$

$\frac{8}{2(A+B)} \le \frac{A+B}{AB}$

$\frac{8}{2(A+B)} \le \frac{1}{A}+\frac{1}{B}$ [Equation 1]

$(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2=x_1y_1+x_2y_2+x_1y_2+x_2y_1-z_1^2-2z_1z_2-z_2^2$

$(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2=(A+B)+x_1y_2+x_2y_1-2z_1z_2$

since $x_1y_1>z_1^2$ and $x_2y_2>z_2^2$, and using the Rearrangement inequality

then $x_1y_1+x_2y_2-z_1z_1-z_2z_2 \le x_1y_2+x_2y_1-z_1z_2-z_1z_2$

$(A+B) \le x_1y_2+x_2y_1-2z_1z_2$

$2(A+B) \le x_1y_1+x_2y_2+x_1y_2+x_2y_1-z_1^2-2z_1z_2-z_2^2$

$2(A+B) \le (x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2$ [Equation 2]

Therefore, we can can use [Equation 2] into [Equation 1] to get:

$\frac{8}{(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2} \le \frac{1}{A}+\frac{1}{B}$

Then, from the values of $A$ and $B$ we get:

$\frac{8}{(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2} \leq \frac{1}{x_1y_1 - z_1^2} + \frac{1}{x_2y_2 - z_2^2}$

With equality at $x_1y_1 - z_1^2=x_2y_2 - z_2^2>0$ and $x_1=x_2, y_1=y_2, z_1=z_2$

~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 $a_i, b_i$, for $i=1,2,...,n$ with $a_i > 0, b_i > 0$

and $\prod_{i=1}^{n-1}a_i-a_n^{n-1}>0, \prod_{i=1}^{n-1}b_i-b_n^{n-1}>0$ the inequality

\[\frac{2^n}{\prod_{i=1}^{n-1}(a_i + b_i) - (a_n + b_n)^{n-1}} \leq \frac{1}{\prod_{i=1}^{n-1}a_i-a_n^{n-1}} + \frac{1}{\prod_{i=1}^{n-1}b_i-b_n^{n-1}}\]is satisfied.

Let $A=\prod_{i=1}^{n-1}a_i-a_n^{n-1}$ and $\prod_{i=1}^{n-1}b_i-b_n^{n-1}>0$

From AM-GM:

$\sqrt{AB} \le \frac{A+B}{2}$ with equality at $A=B$

$4AB \le (A+B)^2$

$\frac{4}{A+B} \le \frac{A+B}{AB}$

$\frac{2^n}{2^{n-2}(A+B)} \le \frac{A+B}{AB}$

$\frac{2^n}{2^{n-2}(A+B)} \le \frac{1}{A}+\frac{1}{B}$ [Equation 3]

Here's the difficult part where I'm skipping steps:

we prove that $2^{n-2}(A+B) \le \prod_{i=1}^{n-1}(a_i + b_i) - (a_n + b_n)^{n-1}$

and replace in [Equation 3] to get:

$\frac{2^n}{\prod_{i=1}^{n-1}(a_i + b_i) - (a_n + b_n)^{n-1}} \le \frac{1}{A}+\frac{1}{B}$

and replace the values of $A$ and $B$ to get:

$\frac{2^n}{\prod_{i=1}^{n-1}(a_i + b_i) - (a_n + b_n)^{n-1}} \leq \frac{1}{\prod_{i=1}^{n-1}a_i-a_n^{n-1}} + \frac{1}{\prod_{i=1}^{n-1}b_i-b_n^{n-1}}$

with equality at $a_i=b_i$ for all $i=1,2,...,n$

Then set $n=3, a_1=x_1, a_2=y_1, a_3=z_1, b_1=x_2, b_2=y_2, b_3=z_3$ and substitute in the generalized inequality to get:

$\frac{8}{(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2} \leq \frac{1}{x_1y_1 - z_1^2} + \frac{1}{x_2y_2 - z_2^2}$

with equality at $x_1=x_2, y_1=y_2, z_1=z_2$

~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 $x_1y_1+x_2y_2-z_1z_1-z_2z_2 \le x_1y_2+x_2y_1-z_1z_2-z_1z_2$ is false. For a counterexample take $x_1 = y_1 = 2, x_2 = y_2 = 1, z_1 = z_2 = 0.5$. The left hand side equals $5 - 0.25 - 0.25$ and the right hand side equals $4 - 0.25 - 0.25$.

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 $y_1 > 0, y_2 > 0$. Also, we can assume $z_1 \ge 0, z_2 \ge 0$. Indeed, if $z_1 < 0$, then $z_1 < -z_1$. It follows $z_1 + z_2 < -z_1 + z_2$, so \[\frac{1}{(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2} < \frac{1}{(x_1 + x_2)(y_1 + y_2) - (-z_1 + z_2)^2}\]

So, if we proved the inequality for positive $z_1$ it is also true for negative $z_1$.

Now consider the function $F$ of three variables $F(x, y, z) = \frac{1}{xy - z^2}$ defined on the domain $x, y > 0, z \ge 0, z^2 < xy$. For simplicity, denote a point $(x, y, z)$ in the domain by $P$. The inequality in the problem can be rewritten as \[F \left( \frac{P_1 + P_2}{2} \right) \le \frac{1}{2}[F(P_1) + F(P_2)]\]

This follows immediately from the inequality expressing the fact that $F(x, y, z)$ is a convex function. (In fact, it is equivalent to the convexity of $F$, but this is not needed for our proof of the given inequality.) Therefore, it is enough to prove that the function $F(x, y, z)$ 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 $P_1 = P_2$, in other words, $x_1 = x_2, y_1 = y_2, z_1 = z_2$, 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
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