prove (or disprove) $\left|x^\top\left(\frac{x}{|x|^\frac{1}{2}}-\frac{y+x}{|y+x|^\frac{1}{2}}\right)\right|\le 3|x||y|^\frac{1}{2}$

Question

As stated in the title, I need help showing (or disproving) $$\left|x^\top\left(\frac{x}{|x|^\frac{1}{2}}-\frac{y+x}{|y+x|^\frac{1}{2}}\right)\right|\le 3|x||y|^\frac{1}{2}$$ for $x, y\in\mathbb{R}^N$.

Here, we take $|\cdot|$ to be the Euclidean norm and we interpret $\frac{x}{|x|^\frac{1}{2}}$ to be $0$ at $x=0$. I have already shown it to be true for scalar x, y but I am having trouble proving (or disproving) it for the vector valued case. I tried several approaches including inducting on the dimension, finding scalars $\tilde{x}, \tilde{y}$ corresponding to $x$ and $y$ (with $|\tilde{x}|<|x|$ and $|\tilde{y}|<|y|$) and upper bounding the vector valued expression by the scalar one, but none of these approaches have really panned out for me. I even wrote a MATLAB script to look for a counter example but cannot find one so far. This problem is relevant to a research problem I am working on so any help would be appreciated. Thanks!

(EDIT: it should be noted that it may possible one could improve the 3 with a smaller constant, but at the moment I can't find an optimal value. However, for my purposes showing the statement with any constant in place of the 3 will suffice)

Answer 1

Since you allow it, here’s a proof with a weaker constant 5 instead of 3.

Let $a = |x|$, $b = \langle x,y\rangle$, and $c = |y|$. If $a \leq 4c$, then we can use Cauchy-Schwarz to bound the quantity by \begin{align*} |x| \cdot \left\lvert \frac{x}{|x|^{1/2}} - \frac{y+x}{|y+x|^{1/2}} \right\rvert \leq |x|\cdot (|x|^{1/2} + |y+x|^{1/2}) \leq a \cdot (2a ^{1/2} + c^{1/2}) \leq a\cdot (2\sqrt 4+1)c^{1/2} = 5a c^{1/2}. \end{align*} We thus assume $a > 4c$. Note that $$ \left\langle x, \frac{x}{|x|^{1/2}} - \frac{y+x}{|y+x|^{1/2}}\right\rangle = a^{3/2} - \frac{b+a^2}{(a^2 + 2b + c^2)^{1/4}} $$ We will show that the second term is convex in $b$. Indeed, the first derivative with respect to $b$ is $$ \frac{(a^2+2b+c^2)^{1/4} - \frac12 (a^2+b) (a^2+2b+c^2)^{-3/4}} {(a^2+2b+c^2)^{2/4}} = (a^2+2b+c^2)^{-1/4} - \frac12 \frac{a^2+b}{(a^2+2b+c^2)^{5/4}} $$ and the second derivative with respect to $b$ is \begin{align*} &-\frac12(a^2+2b+c^2)^{-5/4} - \frac12\frac{(a^2+2b+c^2)^{5/4} - 2\cdot\frac54(a^2+b)(a^2+2b+c^2)^{1/4}}{(a^2+2b+c^2)^{10/4}} \\ =&~ -(a^2+2b+c^2)^{-5/4} + \frac54\frac{a^2+b}{(a^2+2b+c^2)^{9/4}} \\ =&~ -(a^2+2b+c^2)^{-5/4} + \frac54\frac{a^2+b}{(a^2+2b+c^2)^{9/4}} + \frac54\frac{c^2+b}{(a^2+2b+c^2)^{9/4}} - \frac54\frac{c^2+b}{(a^2+2b+c^2)^{9/4}} \\ =&~ \frac14\frac{1}{(a^2+2b+c^2)^{5/4}} - \frac54\frac{c^2+b}{(a^2+2b+c^2)^{9/4}} \\ \end{align*} This is positive as long as $$ a^2+2b+c^2 \geq 5(c^2+b) $$ or $$ a^2-3b+(9/4)c^2 \geq (25/4)c^2 $$ Note that $|b| \leq ac$ by the Cauchy-Schwarz inequality, so this is true if $(a-(3/2)c)^2 \geq (25/4)c^2$, or $a \geq 4c$.

Now, we know that $(b+a^2)/(a^2+2b+c^2)^{1/4}$ is convex, so it is largest or smallest at its endpoints $|b| = ac$ or its critical point. At the end points of $|b| = ac$, this means that $\langle x,y\rangle = |x||y|$, so $x$ and $y$ have to be in the same direction, and this reduces to the 1-dimensional problem, which you have already solved. It remains to check the critical point.

When the derivative is $0$, we have $(a^2+2b+c^2)^{-1/4} = (a^2+b)(a^2+2b+c^2)^{-5/4}/2$, or $b = -(a^2+2c^2)/3$. However, note that we restricted $|b| \leq ac \leq a^2/4$. Thus, this critical point is infeasible, so we dismiss this case and conclude.

Answer 2

Let us prove a stronger inequality: $$\left|x^\top\left(\frac{x}{|x|^\frac{1}{2}}-\frac{y+x}{|y+x|^\frac{1}{2}}\right)\right|\le 2|x||y|^\frac{1}{2}. \tag{1}$$

The non-trivial case is that $x \ne 0$ and $x + y \ne 0$.

By Cauchy-Bunyakovsky-Schwarz inequality, it suffices to prove that $$|x|^2 \cdot \left|\frac{x}{|x|^\frac{1}{2}}-\frac{y+x}{|y+x|^\frac{1}{2}}\right|^2\le 4|x|^2|y|. \tag{2}$$

It suffices to prove that $$ \left|\frac{x}{|x|^\frac{1}{2}}-\frac{y+x}{|y+x|^\frac{1}{2}}\right|^2\le 4|y|$$ or $$|x| + |y + x| - \frac{2(x^\top y + |x|^2)}{|x|^{1/2}|x+y|^{1/2}} \le 4|y|. \tag{3}$$

Using Cauchy-Bunyakovsky-Schwarz inequality, we have $x^\top y \ge -|x|\cdot |y|$. It suffices to prove that $$|x| + |y + x| - \frac{2(-|x|\cdot |y| + |x|^2)}{|x|^{1/2}|x+y|^{1/2}} \le 4|y|. \tag{4}$$

Let $$a := |x|^{1/2}, \quad b := |y|^{1/2}, \quad c := |y + x|^{1/2}.$$ Using Cauchy-Bunyakovsky-Schwarz inequality and $|x + y|^2 = |x|^2 + |y|^2 + 2x^\top y$, we have $$a^4 + b^4 - 2a^2b^2 \le c^4 \le a^4 + b^4 + 2a^2b^2$$ which gives $$\sqrt{|a^2 - b^2|} \le c \le \sqrt{a^2 + b^2}. \tag{5}$$

(4) is written as $$a^2 + c^2 - \frac{2(-a^2b^2 + a^4)}{ac} \le 4b^2. \tag{6}$$

From (6), it suffices to prove that $$-c^3 + (-a^2 + 4b^2)c + 2a^3 - 2ab^2 \ge 0. \tag{7}$$

From (5), using $c^2 \le a^2 + b^2$, it suffices to prove that $$-c\cdot (a^2 + b^2) + (-a^2 + 4b^2)c + 2a^3 - 2ab^2 \ge 0$$ or $$2a(a^2 - b^2) - (2a^2 - 3b^2)c \ge 0. \tag{8}$$

If $2a^2 - 3b^2 \le 0$, from (5), it suffices to prove that $$2a(a^2 - b^2) - (2a^2 - 3b^2)\sqrt{|a^2 - b^2|} \ge 0$$ which is true (easy).

If $2a^2 - 3b^2 > 0$, from (5), it suffices to prove that $$2a(a^2 - b^2) - (2a^2 - 3b^2)\sqrt{a^2 + b^2} \ge 0$$ which is true (easy).

We are done.

Answer 3

Remarks: I think that $\sqrt 2$ is the best constant. It is not proved yet.

Problem. Let $x, y\in \mathbb{R}^n$ be vectors with $x \ne 0$ and $x + y \ne 0$. Prove that $$\left|x^\top\left(\frac{x}{|x|^\frac{1}{2}}-\frac{y+x}{|y+x|^\frac{1}{2}}\right)\right|\le \sqrt 2\,|x||y|^\frac{1}{2}.$$ Here, $|x| = \sqrt{x^\mathsf{T}x}$ is the Euclidean norm.

proof.

By Cauchy-Bunyakovsky-Schwarz inequality, it suffices to prove that $$|x|^2 \cdot \left|\frac{x}{|x|^\frac{1}{2}}-\frac{y+x}{|y+x|^\frac{1}{2}}\right|^2\le 2|x|^2|y|.$$

It suffices to prove that $$ \left|\frac{x}{|x|^\frac{1}{2}}-\frac{y+x}{|y+x|^\frac{1}{2}}\right|^2\le 2|y|$$ or $$|x| + |y + x| - \frac{2(x^\top y + |x|^2)}{|x|^{1/2}|x+y|^{1/2}} \le 2|y|$$ or $$|x| + \sqrt{|x|^2 + |y|^2 + 2x^\mathsf{T} y} - \frac{2(x^\top y + |x|^2)}{|x|^{1/2}\sqrt[4]{|x|^2 + |y|^2 + 2x^\mathsf{T} y}} \le 2|y|. \tag{1}$$

WLOG, assume that $|x| = 1$. Let $b: = |y|$ and $u := x^\mathsf{T}y$. Then $- b \le u \le b$.

We need to prove that $$1 + \sqrt{1 + b^2 + 2u} - \frac{2(u + 1)}{\sqrt[4]{1 + b^2 + 2u}} \le 2b. \tag{2}$$

If $b = 1$, true (easy).

If $b \ne 1$, letting $v = \sqrt[4]{1 + b^2 + 2u}$, we have $\sqrt{|b - 1|} \le v \le \sqrt{b + 1}$. It suffices to prove that $$1 + v^2 - \frac{v^4 + 1 - b^2}{v} \le 2b$$ which is true (not difficult).

We are done.