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

Latest revision as of 13:50, 1 August 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.

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 $F(x, y, z)$ is the 3x3 matrix $H$ formed by the second derivatives of $F$ :

$\frac{\partial^2 F}{\partial x^2} \hspace{20pt} \frac{\partial^2 F}{\partial x \partial y} \hspace{20pt} \frac{\partial^2 F}{\partial x \partial z} \\ \\ \\ \frac{\partial^2 F}{\partial y \partial x} \hspace{20pt} \frac{\partial^2 F}{\partial y^2} \hspace{20pt} \frac{\partial^2 F}{\partial y \partial z} \\ \\ \\ \frac{\partial^2 F}{\partial z \partial x} \hspace{20pt} \frac{\partial^2 F}{\partial z \partial y} \hspace{20pt} \frac{\partial^2 F}{\partial z^2}$

$H$ is positive definite if for any 3x1 vector $v$ we have $v^t H v \ge 0$ . ( $v^t$ is the transpose of $v$ , a matrix which is 1x3.) $H$ is strictly positive definite if $v^t H v > 0$ unless $v = 0$ .

Explicitly, we need to show that if $a, b, c$ are three numbers, then

$a^2 \frac{\partial^2 F}{\partial x^2} + ab \frac{\partial^2 F}{\partial x \partial y} + ac \frac{\partial^2 F}{\partial x \partial z} + ab \frac{\partial^2 F}{\partial y \partial x} + b^2 \frac{\partial^2 F}{\partial y^2} + bc \frac{\partial^2 F}{\partial y \partial z} + ac \frac{\partial^2 F}{\partial z \partial x} + bc \frac{\partial^2 F}{\partial z \partial y} + c^2 \frac{\partial^2 F}{\partial z^2} \ge 0$

and that the expression is $= 0$ only when $a = b = c = 0$ .

Before embarking on the computations proving the inequality expressing the positive definiteness of the Hessian of $F$ , let us show that the domain of $F$ is convex. Let $P_1 = (x_1, y_1, z_1)$ and $P_2 = (x_2, y_2, z_2)$ be two points in the domain of $F$ . This means that $x_1, y_1, x_2, y_2 > 0$ , $z_1, z_2 \ge 0$ , $z_1^2 < x_1 y_1$ and $z_2^2 < x_2 y_2$ . We need to verify that the same is true for $P = t P_1 + (1 - t) P_2$ when $0 < t < 1$ . The only thing which is not obvious is $[t z_1 + (1 - t) z_2]^2 < [t x_1 + (1 - t) x_2] [t y_1 + (1 - t) y_2]$ . To see this, work out the multiplications, use what we already know about $z_1^2$ and $z_2^2$ , simplify by $t(1-t)$ and we are left with showing that $2 z_1 z_2 < x_1 y_2 + x_2 y_1$ . This can be seen as follows: $2 z_1 z_2 < 2 \sqrt{x_1 y_1} \sqrt{x_2 y_2} = 2 \sqrt{(x_1 y_2) (x_2 y_1)} \le x_1 y_2 + x_2 y_1$ .

Now work towards proving the inequality expressing that the Hessian of $F$ is positive definite. Compute all the second derivatives of $F$ ; notice that they all have $(xy - z^2)^3$ 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 $\frac{\partial^2 F}{\partial x \partial y} = \frac{\partial^2 F}{\partial y \partial x}$ ).)

We get (as the thing we have to prove) that

$2a^2y^2 + 2ab(xy + z^2) - 8acyz + 2b^2x^2 - 8bcxz + c^2(2xy + 6z^2) \ge 0$ .

If $a = b = 0$ the expression equals $0$ if and only if $c = 0$ as well. So, assume that at least one of $a, b$ is $\ne 0$ . We can assume $a \ne 0$ . We will show that in this case, the expression is $> 0$ .

Simplify by $2$ , and rearrange the terms so we can think of the expression on the left as a polynomial in $c$ :

$(xy + 3z^2)c^2 - 4(xb + ya)zc + [x^2b^2 + y^2a^2 + (xy + z^2)ab] > 0$ .

Notice that this is a polynomial (in $c$ ) of degree $2$ of the type $AX^2 + BX + C$ , whose leading coefficient is $> 0$ . To show that its values are always $> 0$ we need to show that its discriminant is $< 0$ , in other words, $B^2 - 4AC < 0$ .

Therefore, we need to show that

$16[(xb + ya)z]^2 - 4(xy + 3z^2)[x^2b^2 + y^2a^2 + (xy + 3z^2)ab] < 0$

After factoring out $4$ , and working out all multiplications, this becomes

$-x^3yb^2 + x^2z^2b^2 - xy^3a^2 + y^2z^2a^2 - x^2y^2ab + 4xyz^2ab - 3z^4ab < 0$

After some rearranging of terms, grouping, and factoring, this becomes

$-(xy - z^2)[x^2b^2 + (xy - 3z^2)ab + y^2a^2] < 0$

Since $x^2 < xy$ , we need to show

$x^2b^2 + (xy - 3z^2)ab + y^2a^2 > 0$

Now, view the expression on the left hand side as a polynomial of degree $2$ in $b$ , whose leading coefficient is $> 0$ . To conclude that its values are $> 0$ we need to show that its discriminant is $< 0$ . That is, we need to show that

$[(xy - 3z^2)a]^2 - 4x^2y^2a^2 < 0$

Work out the multiplications, and regroup terms to get

$-3a^2(xy - z^2)(xy + 3z^2) < 0$ .

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.

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

Latest revision as of 13:50, 1 August 2024

Contents

Problem

Solution

Generalization and Idea for a Solution

Remarks (added by pf02, July 2024)

Solution

See Also

@@ Line 1: / Line 1: @@
 ==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, 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</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==
@@ Line 8: / Line 11: @@
 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>
@@ Line 20: / Line 23: @@
 <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>
-{{solution}}
+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
+==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 <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
+==Remarks (added by pf02, July 2024)==
+. 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>.
+. 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.
+. 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} =
+\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}}
 == See Also == {{IMO box|year=1969|num-b=5|after=Last Question}}