1969 IMO Problems/Problem 6

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: $2 z_1 z_2 \le 2 |z_1| |z_2| < 2 \sqrt{x_1 y_1} \sqrt{x_2 y_2} = 2 \sqrt{(x_1 y_2) (x_2 y_1)} \le x_1 y_2 + x_2 y_1$ .

\\ [TO BE CONTINUED. SAVING, SO THAT I DON'T LOSE 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.

1969 IMO (Problems) • Resources
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1969 IMO Problems/Problem 6

Contents

Problem

Solution

Generalization and Idea for a Solution

Remarks (added by pf02, July 2024)

Solution

See Also