Kvant Math Problem 543
The expression
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Problem
Let $$\rho(x,y)=\dfrac{|x-y|}{\sqrt{1+x^2}\sqrt{1+\smash{y}\vphantom{x}^2}}.$$ Prove that for any real numbers $a$, $b$, $c$ the inequality $$\rho(a,c)\le\rho(a,b)+\rho(b,c).$$ holds.
Andrzej Paszkiewicz (Warsaw)
Exploration
The expression
$$\rho(x,y)=\frac{|x-y|}{\sqrt{1+x^2}\sqrt{1+y^2}}$$
resembles a Euclidean distance after a trigonometric substitution. If we put
$$x=\tan \alpha,\qquad y=\tan \beta,$$
then
$$x-y=\frac{\sin(\alpha-\beta)}{\cos\alpha\cos\beta},$$
and
$$\sqrt{1+x^2}=\sec\alpha,\qquad \sqrt{1+y^2}=\sec\beta.$$
Hence
$$\rho(x,y)=|\sin(\alpha-\beta)|.$$
The problem then becomes
$$|\sin(\alpha-\gamma)| \le |\sin(\alpha-\beta)|+|\sin(\beta-\gamma)|.$$
This looks much closer to a triangle inequality. The quantity $|\sin(u-v)|$ is the Euclidean distance between the points $(\cos u,\sin u)$ and $(\cos v,\sin v)$ on the unit circle, divided by $2$, because
$$(\cos u-\cos v)^2+(\sin u-\sin v)^2 =4\sin^2\frac{u-v}{2}.$$
That gives the chord length $2|\sin\frac{u-v}{2}|$, not $|\sin(u-v)|$, so this route does not immediately produce the desired expression.
A better observation is that
$$|\sin(\alpha-\beta)| = |\sin\alpha\cos\beta-\cos\alpha\sin\beta|.$$
This is the absolute value of the determinant of the unit vectors
$$u_\alpha=(\cos\alpha,\sin\alpha),\qquad u_\beta=(\cos\beta,\sin\beta).$$
For vectors $u,v,w$ in the plane,
$$\det(u,w)=\det(u,v)+\det(v,w).$$
Taking absolute values yields
$$|\det(u,w)| \le |\det(u,v)|+|\det(v,w)|.$$
This is exactly the desired inequality after the tangent substitution.
The step most likely to hide an error is the passage from real numbers $a,b,c$ to angles. One must choose angles with $\tan\alpha=a$, $\tan\beta=b$, $\tan\gamma=c$, and verify carefully that $\rho(x,y)=|\sin(\alpha-\beta)|$ without sign mistakes.
Problem Understanding
We must prove that the function
$$\rho(x,y)=\frac{|x-y|}{\sqrt{1+x^2}\sqrt{1+y^2}}$$
satisfies the triangle inequality
$$\rho(a,c)\le \rho(a,b)+\rho(b,c)$$
for all real numbers $a,b,c$.
This is a Type B problem. The statement is already given, and the task is to prove it.
The core difficulty is to recognize the geometric meaning of $\rho$. After the substitution $x=\tan\theta$, the expression becomes the absolute value of a determinant of two unit vectors. The triangle inequality then follows from the linearity of the determinant.
Proof Architecture
Lemma 1. For any real numbers $x,y$, if $\alpha,\beta\in(-\pi/2,\pi/2)$ satisfy $x=\tan\alpha$ and $y=\tan\beta$, then
$$\rho(x,y)=|\sin(\alpha-\beta)|.$$
This follows from the tangent subtraction formula and $\sqrt{1+\tan^2 t}=\sec t$.
Lemma 2. If $u_\theta=(\cos\theta,\sin\theta)$, then
$$\det(u_\alpha,u_\beta)=\sin(\beta-\alpha).$$
This is a direct computation.
Lemma 3. For any planar vectors $u,v,w$,
$$\det(u,w)=\det(u,v)+\det(v,w).$$
This follows from bilinearity and antisymmetry of the determinant.
Lemma 4. For any planar vectors $u,v,w$,
$$|\det(u,w)| \le |\det(u,v)|+|\det(v,w)|.$$
Apply the ordinary triangle inequality to Lemma 3.
The hardest point is Lemma 1, because the entire reduction depends on the exact identity $\rho(x,y)=|\sin(\alpha-\beta)|$.
Solution
Choose angles
$$\alpha,\beta,\gamma\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$$
such that
$$a=\tan\alpha,\qquad b=\tan\beta,\qquad c=\tan\gamma .$$
We first express $\rho$ in terms of these angles. Using
$$\tan\alpha-\tan\beta = \frac{\sin(\alpha-\beta)}{\cos\alpha\cos\beta},$$
and
$$\sqrt{1+\tan^2\alpha}=\sec\alpha=\frac1{\cos\alpha},$$
we obtain
$$\rho(a,b) = \frac{\left|\dfrac{\sin(\alpha-\beta)} {\cos\alpha\cos\beta}\right|} {\sec\alpha,\sec\beta} = |\sin(\alpha-\beta)|.$$
Similarly,
$$\rho(b,c)=|\sin(\beta-\gamma)|, \qquad \rho(a,c)=|\sin(\alpha-\gamma)|.$$
Now let
$$u_\theta=(\cos\theta,\sin\theta).$$
A direct computation gives
$$\det(u_\alpha,u_\beta) = \cos\alpha\sin\beta-\sin\alpha\cos\beta = \sin(\beta-\alpha).$$
Hence
$$\rho(a,b)=|\det(u_\alpha,u_\beta)|.$$
Analogous formulas hold for the other pairs.
Using bilinearity of the determinant,
$$\det(u_\alpha,u_\gamma) = \det(u_\alpha,u_\beta) + \det(u_\beta,u_\gamma).$$
Taking absolute values and applying the ordinary triangle inequality,
$$|\det(u_\alpha,u_\gamma)| \le |\det(u_\alpha,u_\beta)| + |\det(u_\beta,u_\gamma)|.$$
Replacing each determinant by the corresponding $\rho$-value yields
$$\rho(a,c) \le \rho(a,b)+\rho(b,c).$$
This completes the proof.
∎
Verification of Key Steps
The identity
$$\rho(a,b)=|\sin(\alpha-\beta)|$$
must be checked carefully. Starting from
$$a=\tan\alpha,\qquad b=\tan\beta,$$
we have
$$a-b = \frac{\sin(\alpha-\beta)}{\cos\alpha\cos\beta}.$$
Since $\alpha,\beta\in(-\pi/2,\pi/2)$, both cosines are positive, and
$$\sqrt{1+a^2}\sqrt{1+b^2} = \frac1{\cos\alpha\cos\beta}.$$
Multiplying gives exactly
$$\rho(a,b)=|\sin(\alpha-\beta)|.$$
The determinant identity also deserves an independent check:
$$\det(u_\alpha,u_\gamma) = \det(u_\alpha,u_\beta) + \det(u_\beta,u_\gamma).$$
Writing $u_\gamma=u_\beta+(u_\gamma-u_\beta)$,
$$\det(u_\alpha,u_\gamma) = \det(u_\alpha,u_\beta) + \det(u_\alpha,u_\gamma-u_\beta).$$
Since
$$\det(u_\alpha,u_\gamma-u_\beta) = \det(u_\alpha,u_\gamma)-\det(u_\alpha,u_\beta),$$
the identity reduces to
$$\det(u_\alpha,u_\gamma) = \det(u_\alpha,u_\beta)+\det(u_\beta,u_\gamma),$$
which is equivalent to
$$\det(u_\alpha-u_\beta,u_\gamma-u_\beta)=0.$$
Expanding confirms the equality.
A common mistake is to identify $|\sin(\alpha-\beta)|$ with the distance between two points on the unit circle. The actual chord length is
$$2\left|\sin\frac{\alpha-\beta}{2}\right|,$$
so that interpretation does not directly yield the required triangle inequality.
Alternative Approaches
After the substitution $a=\tan\alpha$, $b=\tan\beta$, $c=\tan\gamma$, the problem becomes
$$|\sin(\alpha-\gamma)| \le |\sin(\alpha-\beta)|+|\sin(\beta-\gamma)|.$$
One may then write
$$\sin(\alpha-\gamma) = \sin(\alpha-\beta)\cos(\beta-\gamma) + \cos(\alpha-\beta)\sin(\beta-\gamma)$$
and apply the triangle inequality:
$$|\sin(\alpha-\gamma)| \le |\sin(\alpha-\beta)|,|\cos(\beta-\gamma)| + |\cos(\alpha-\beta)|,|\sin(\beta-\gamma)|.$$
Since both cosine factors have absolute value at most $1$,
$$|\sin(\alpha-\gamma)| \le |\sin(\alpha-\beta)| + |\sin(\beta-\gamma)|.$$
This proof is shorter. The determinant approach is preferable because it reveals the geometric structure behind $\rho$: after the tangent parametrization, $\rho$ becomes the absolute value of the oriented area of the parallelogram spanned by two unit vectors. The triangle inequality then becomes an immediate consequence of linearity.