How to prove this inequality $\left|x\sin{\frac{1}{x}}-y\sin{\frac{1}{y}}\right|<2\sqrt{|x-y|}$?

Question

For any real numbers $x,y\neq 0$,show that $$\left|x\sin{\dfrac{1}{x}}-y\sin{\dfrac{1}{y}}\right|<2\sqrt{|x-y|}$$ I found this problem when I dealt with this problem. But I can't prove it. Maybe the constant $2$ on the right hand side can be replaced by the better constant $\sqrt{2}$?

Thank you.

Answer 1

We'll first assume that $0 \le x < y$.

I need three kinds of estimates here.

1. $\left| x \sin \frac{1}{x} - y \sin \frac{1}{y} \right| \le x + y$
2. $\left| x \sin \frac{1}{x} - y \sin \frac{1}{y} \right| \le 2$
3. $\left| x \sin \frac{1}{x} - y \sin \frac{1}{y} \right| \le \frac{y}{x} - 1$

Estimates 1 and 2 are trivial: $\left| x \sin \frac{1}{x} \right| \le \min(x, 1)$

Estimate 3 is proved as follows.

• First we show that $\left| (x \sin \frac{1}{x})^\prime \right| \le \frac{1}{x}$. Indeed, $(x \sin \frac{1}{x})^{\prime\prime} = - \frac{1}{x^3} \sin \frac{1}{x}$, so local maxima and minima of $(x \sin \frac{1}{x})^\prime$ are located at $\frac{1}{\pi n}, n \in \mathbb{N}$, and its values there are $\frac{(-1)^n}{\pi n}$, so that $\sup_{z \ge x} \left| (z \sin \frac{1}{z})^\prime \right| \le \frac{1}{\pi n} \le \frac{1}{x}$, where $\frac{1}{\pi n}$ is the smallest one in $[x,+\infty)$.
• Now $\left| x \sin \frac{1}{x} - y \sin \frac{1}{y} \right| \le \intop_x^y \frac{1}{z} dz = \log \frac{y}{x} \le \frac{y}{x} - 1$

Now let's find the three regions corresponding to:

1. $x + y \le C \sqrt{y - x}$
2. $2 \le C \sqrt{y - x}$
3. $\frac{y}{x} - 1 \le C \sqrt{y - x}$

and choose the constant $C$ in such a way that these regions cover $\{0 < x < y\}$. Clearly, 2 is bounded by a line, 1 and 3 are bounded by parabolas. Miraculously, $C = 2$ is the unique value when the three boundaries intersect at a single point, namely $(x,y) = (1/2, 3/2)$...

Anyway, let's rewrite our regions for $C = 2$:

1. $y-x \ge \frac{1}{4} (x+y)^2$
2. $y-x \ge 1$
3. $y-x \le 4 x^2$

So to cover the whole space we have to prove $y-x \ge \frac{1}{4} (x+y)^2$ on $\{4 x^2 \le y-x \le 1\}$, for which it is sufficient to consider just the two boundary cases, namely $y-x = 1$ and $y-x = 4 x^2$, since a quadratic polynomial with positive leading term attains maximum on the boundary of a segment. And these cases are easy to verify. Indeed, the inequality in terms of $x$ and $z := y-x$ looks like $x^2 + xz + \frac{1}{4} z^2 \le z$; for $z=1$ it's equivalent to $(x + \frac{1}{2})^2 \le 1$, which is true, since $4 x^2 \le 1$; for $z = 4 x^2$ it's equivalent to $3 x^2 \ge 4 x^3 + 4 x^4$, which follows once again from $4 x^2 \le 1$ (since $x^2 \ge 2 x^3$ and $x^2 \ge 4 x^4$).

Now the case $x < 0 < y$ is even simpler. We only need analogs of estimates 1 and 2:

1. $\left| x \sin \frac{1}{x} - y \sin \frac{1}{y} \right| \le |x| + |y| \le C \sqrt{|x| + |y|}$ whenever $|x| + |y| \le C^2 = 4$
2. $\left| x \sin \frac{1}{x} - y \sin \frac{1}{y} \right| \le 2 \le C \sqrt{|x| + |y|}$ whenever $|x| + |y| \ge (\frac{2}{C})^2 = 1$.

Answer 2

This is a partial alternative solution for the case $\left|\frac{1}{x}-\frac{1}{y}\right|\leq 1$.

We can clearly assume that both $x$ and $y$ are positive, then, through the substitutions $x\to 1/x,y\to 1/y$, prove that: $$ \forall x,y>0,x\neq y,\quad (y\sin x-x\sin y)^2 < 4xy|y-x|. $$ The LHS can be written as: $$ \left((y-x)\sin x + x\,(\sin x-\sin y)\right)^2, $$ that, by the Cauchy-Schwarz inequality, satisfies: $$ \left((y-x)\sin x + x\,(\sin x-\sin y)\right)^2 \leq (x^2+\sin^2 x)\left((y-x)^2+(\sin y-\sin x)^2\right), $$ and the RHS is less than $4x^2(y-x)^2$, since $\sin x$ is a $1$-Lipschitz function.

By exchanging $x$ and $y$, we have: $$ (y\sin x-x\sin y)^2 < \min\left(4x^2(y-x)^2,4y^2(y-x)^2\right)<4xy(y-x)^2,$$ so the inequality is clearly true if $|y-x|\leq 1$.