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Difference between revisions of "1969 IMO Problems/Problem 6"

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(→‎Solution 2)
 
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From AM-GM:
 
From AM-GM:
  
−
<math>\sqrt{AB} \le \frac{A+B}{2}</math>
+
<math>\sqrt{AB} \le \frac{A+B}{2}</math> with equality at <math>A=B</math>
  
 
<math>4AB \le (A+B)^2</math>
 
<math>4AB \le (A+B)^2</math>
  
−
<math>\frac{4}{A+B} \le \frac{A+B}{2}</math>
+
<math>\frac{4}{A+B} \le \frac{A+B}{AB}</math>
  
 +
<math>\frac{8}{2(A+B)} \le \frac{A+B}{AB}</math>
  
−
{{solution}}
+
<math>\frac{8}{2(A+B)} \le \frac{1}{A}+\frac{1}{B}</math> [Equation 1]
 +
 
 +
<math>(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</math>
 +
 
 +
<math>(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2=(A+B)+x_1y_2+x_2y_1-2z_1z_2</math>
 +
 
 +
since <math>x_1y_1>z_1^2</math> and <math>x_2y_2>z_2^2</math>, and using the Rearrangement inequality
 +
 
 +
then <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>
 +
 
 +
<math>(A+B) \le  x_1y_2+x_2y_1-2z_1z_2</math>
 +
 
 +
<math>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</math>
 +
 
 +
<math>2(A+B) \le (x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2</math> [Equation 2]
 +
 
 +
Therefore, we can can use [Equation 2] into [Equation 1] to get:
 +
 
 +
<math>\frac{8}{(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2} \le \frac{1}{A}+\frac{1}{B}</math>
 +
 
 +
Then, from the values of <math>A</math> and <math>B</math> we get:
 +
 
 +
<math>\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}</math>
 +
 
 +
With equality at <math>x_1y_1 - z_1^2=x_2y_2 - z_2^2>0</math> and <math>x_1=x_2, y_1=y_2, z_1=z_2</math>
 +
 
 +
~Tomas Diaz. orders@tomasdiaz.com
 +
 
 +
==Solution 2==
 +
 
 +
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 <math>a_i, b_i</math>, for <math>i=1,2,...,n</math> with <math>a_i > 0, b_i > 0</math>
 +
 
 +
and <math>\prod_{i=1}^{n-1}a_i-a_n^{n-1}>0, \prod_{i=1}^{n-1}b_i-b_n^{n-1}>0</math> the inequality
 +
 
 +
<cmath>\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}}</cmath>is satisfied.
 +
 
 +
Let <math>A=\prod_{i=1}^{n-1}a_i-a_n^{n-1}</math> and <math>\prod_{i=1}^{n-1}b_i-b_n^{n-1}>0</math>
 +
 
 +
From AM-GM:
 +
 
 +
<math>\sqrt{AB} \le \frac{A+B}{2}</math> with equality at <math>A=B</math>
 +
 
 +
<math>4AB \le (A+B)^2</math>
 +
 
 +
<math>\frac{4}{A+B} \le \frac{A+B}{AB}</math>
 +
 
 +
<math>\frac{2^n}{2^{n-2}(A+B)} \le \frac{A+B}{AB}</math>
 +
 
 +
<math>\frac{2^n}{2^{n-2}(A+B)} \le \frac{1}{A}+\frac{1}{B}</math> [Equation 3]
 +
 
 +
Here's the difficult part where I'm skipping steps:
 +
 
 +
we prove that <math>2^{n-2}(A+B) \le \prod_{i=1}^{n-1}(a_i + b_i) - (a_n + b_n)^{n-1}</math>
 +
 
 +
and replace in [Equation 3] to get:
 +
 
 +
<math>\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}</math>
 +
 
 +
and replace the values of <math>A</math> and <math>B</math> to get:
 +
 
 +
<math>\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}}</math>
 +
 
 +
with equality at <math>a_i=b_i</math> for all <math>i=1,2,...,n</math>
 +
 
 +
Then set <math>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</math> and substitute in the generalized inequality to get:
 +
 
 +
<math>\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}</math>
 +
 
 +
with equality at <math>x_1=x_2, y_1=y_2, z_1=z_2</math>
 +
 
 +
~Tomas Diaz. orders@tomasdiaz.com
 +
 
 +
{{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}}

Latest revision as of 04:21, 19 November 2023

Problem

Prove that for all real numbers $x_1, x_2, y_1, y_2, z_1, z_2$, with $x_1 > 0, x_2 > 0, y_1 > 0, y_2 > 0, z_1 > 0, z_2 > 0, x_1y_1 - z_1^2 > 0, x_2y_2 - z_2^2 > 0$, 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

Solution 2

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

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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