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, | + | 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== | ||
Line 44: | Line 47: | ||
~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 66: | Line 69: | ||
<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{A+B}{AB}</math> | ||
− | <math>\frac{2^n}{2^{n-2}(A+B)} \le \frac{1}{A}+\frac{1}{B}</math> [Equation | + | <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: | Here's the difficult part where I'm skipping steps: | ||
− | we prove that | + | 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 | ||
+ | |||
+ | ==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 | ||
+ | <cmath>F(x, y, z) = \frac{1}{xy - z^2}</cmath> | ||
+ | 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 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. | ||
+ | |||
+ | 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 <math>F(x, y, z)</math> is the 3x3 matrix <math>H</math> formed | ||
+ | by the second derivatives of <math>F</math>: | ||
+ | |||
+ | <math> | ||
+ | \frac{\partial^2 F}{\partial x^2} \hspace{20pt} | ||
+ | \frac{\partial^2 F}{\partial x \partial y} \hspace{20pt} | ||
+ | \frac{\partial^2 F}{\partial x \partial z} | ||
+ | \\ \\ \\ | ||
+ | \frac{\partial^2 F}{\partial y \partial x} \hspace{20pt} | ||
+ | \frac{\partial^2 F}{\partial y^2} \hspace{20pt} | ||
+ | \frac{\partial^2 F}{\partial y \partial z} | ||
+ | \\ \\ \\ | ||
+ | \frac{\partial^2 F}{\partial z \partial x} \hspace{20pt} | ||
+ | \frac{\partial^2 F}{\partial z \partial y} \hspace{20pt} | ||
+ | \frac{\partial^2 F}{\partial z^2} | ||
+ | </math> | ||
+ | |||
+ | <math>H</math> is positive definite if for any 3x1 vector <math>v</math> we have | ||
+ | <math>v^t H v \ge 0</math>. (<math>v^t</math> is the transpose of <math>v</math>, a matrix which is 1x3.) | ||
+ | <math>H</math> is strictly positive definite if <math>v^t H v > 0</math> unless <math>v = 0</math>. | ||
+ | |||
+ | Explicitly, we need to show that if <math>a, b, c</math> are three numbers, | ||
+ | then | ||
+ | |||
+ | <math> | ||
+ | a^2 \frac{\partial^2 F}{\partial x^2} + | ||
+ | ab \frac{\partial^2 F}{\partial x \partial y} + | ||
+ | ac \frac{\partial^2 F}{\partial x \partial z} + | ||
+ | ab \frac{\partial^2 F}{\partial y \partial x} + | ||
+ | b^2 \frac{\partial^2 F}{\partial y^2} + | ||
+ | bc \frac{\partial^2 F}{\partial y \partial z} + | ||
+ | ac \frac{\partial^2 F}{\partial z \partial x} + | ||
+ | bc \frac{\partial^2 F}{\partial z \partial y} + | ||
+ | c^2 \frac{\partial^2 F}{\partial z^2} \ge 0 | ||
+ | </math> | ||
+ | |||
+ | and that the expression is <math>= 0</math> only when <math>a = b = c = 0</math>. | ||
+ | |||
+ | Before embarking on the computations proving the inequality | ||
+ | expressing the positive definiteness of the Hessian of <math>F</math>, | ||
+ | let us show that the domain of <math>F</math> is convex. Let | ||
+ | <math>P_1 = (x_1, y_1, z_1)</math> and <math>P_2 = (x_2, y_2, z_2)</math> be two | ||
+ | points in the domain of <math>F</math>. This means that <math>x_1, y_1, x_2, y_2 > 0</math>, | ||
+ | <math>z_1, z_2 \ge 0</math>, <math>z_1^2 < x_1 y_1</math> and <math>z_2^2 < x_2 y_2</math>. We need | ||
+ | to verify that the same is true for <math>P = t P_1 + (1 - t) P_2</math> when | ||
+ | <math>0 < t < 1</math>. The only thing which is not obvious is | ||
+ | <math>[t z_1 + (1 - t) z_2]^2 < [t x_1 + (1 - t) x_2] [t y_1 + (1 - t) y_2]</math>. | ||
+ | To see this, work out the multiplications, use what we already know about | ||
+ | <math>z_1^2</math> and <math>z_2^2</math>, simplify by <math>t(1-t)</math> and we are left with showing | ||
+ | that <math>2 z_1 z_2 < x_1 y_2 + x_2 y_1</math>. This can be seen as follows: | ||
+ | <math>2 z_1 z_2 < 2 \sqrt{x_1 y_1} \sqrt{x_2 y_2} = | ||
+ | 2 \sqrt{(x_1 y_2) (x_2 y_1)} \le x_1 y_2 + x_2 y_1</math>. | ||
+ | |||
+ | Now work towards proving the inequality expressing that the Hessian | ||
+ | of <math>F</math> is positive definite. Compute all the second derivatives of | ||
+ | <math>F</math>; notice that they all have <math>(xy - z^2)^3</math> 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 | ||
+ | <math>\frac{\partial^2 F}{\partial x \partial y} = | ||
+ | \frac{\partial^2 F}{\partial y \partial x}</math>).) | ||
+ | |||
+ | We get (as the thing we have to prove) that | ||
+ | |||
+ | <math>2a^2y^2 + 2ab(xy + z^2) - 8acyz + 2b^2x^2 - 8bcxz + c^2(2xy + 6z^2) \ge 0</math>. | ||
+ | |||
+ | If <math>a = b = 0</math> the expression equals <math>0</math> if and only if <math>c = 0</math> as well. | ||
+ | So, assume that at least one of <math>a, b</math> is <math>\ne 0</math>. We can assume <math>a \ne 0</math>. | ||
+ | We will show that in this case, the expression is <math>> 0</math>. | ||
+ | |||
+ | Simplify by <math>2</math>, and rearrange the terms so we can think of the expression | ||
+ | on the left as a polynomial in <math>c</math>: | ||
+ | |||
+ | <math>(xy + 3z^2)c^2 - 4(xb + ya)zc + [x^2b^2 + y^2a^2 + (xy + z^2)ab] > 0</math>. | ||
+ | |||
+ | Notice that this is a polynomial (in <math>c</math>) of degree <math>2</math> of the type | ||
+ | <math>AX^2 + BX + C</math>, whose leading coefficient is <math>> 0</math>. To show that its | ||
+ | values are always <math>> 0</math> we need to show that its discriminant is <math>< 0</math>, | ||
+ | in other words, <math>B^2 - 4AC < 0</math>. | ||
+ | |||
+ | Therefore, we need to show that | ||
+ | |||
+ | <math>16[(xb + ya)z]^2 - 4(xy + 3z^2)[x^2b^2 + y^2a^2 + (xy + 3z^2)ab] < 0</math> | ||
+ | |||
+ | After factoring out <math>4</math>, and working out all multiplications, this becomes | ||
+ | |||
+ | <math>-x^3yb^2 + x^2z^2b^2 - xy^3a^2 + y^2z^2a^2 - x^2y^2ab + 4xyz^2ab - 3z^4ab < 0</math> | ||
+ | |||
+ | After some rearranging of terms, grouping, and factoring, this becomes | ||
+ | |||
+ | <math>-(xy - z^2)[x^2b^2 + (xy - 3z^2)ab + y^2a^2] < 0</math> | ||
+ | |||
+ | Since <math>x^2 < xy</math>, we need to show | ||
+ | |||
+ | <math>x^2b^2 + (xy - 3z^2)ab + y^2a^2 > 0</math> | ||
+ | |||
+ | Now, view the expression on the left hand side as a polynomial of degree | ||
+ | <math>2</math> in <math>b</math>, whose leading coefficient is <math>> 0</math>. To conclude that its | ||
+ | values are <math>> 0</math> we need to show that its discriminant is <math>< 0</math>. That | ||
+ | is, we need to show that | ||
+ | |||
+ | <math>[(xy - 3z^2)a]^2 - 4x^2y^2a^2 < 0</math> | ||
+ | |||
+ | Work out the multiplications, and regroup terms to get | ||
+ | |||
+ | <math>-3a^2(xy - z^2)(xy + 3z^2) < 0</math>. | ||
+ | |||
+ | This is clearly true, which concludes the proof. | ||
+ | |||
+ | [Solution by pf02, July 2024] | ||
{{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}} |
Latest revision as of 14:50, 1 August 2024
Contents
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 |