Inequality $|a|\langle x+a\rangle^{-1} \leq \langle x \rangle$ for all $x,a \in \mathbb R^d$ with strange tangency behavior

Question

While working on a problem in harmonic analysis, attempting to apply DCT, we stumbled across the following interesting inequality: $|a|\langle x+a\rangle^{-1} \leq \langle x \rangle$ for all $x,a \in \mathbb R^d$, where $\langle x\rangle$ is the Japanese bracket of $x$, defined to be $\sqrt{1+|x|^2}$. Writing everything in coordinates, this is the inequality $$(a_1^2+\ldots a_n^2) \leq (1+(x_1+a_1)^2 + \ldots + (x_n+a_n)^2)(1+x_1^2+\ldots + x_n^2).$$ Looking at the graphs on Desmos the graphs look tangent! However, for small values of $a$, they are clearly not tangent, so intuitively one would expect them not to become tangent suddenly for larger values of $a$. Indeed for $a<2$, the curves do not touch, and for $a=2$, they do touch (seemingly tangently) at $x=-1$.

Trying $a=3$, they do touch (again seemingly tangently) at $x=-1-\varphi$ where $\varphi$ is the golden ratio (I recognized this by seeing on Desmos they meet at $x=-2.618$, and miraculously it's the actual golden ratio). The other tangent point is $x=\frac{3-\sqrt 5}2$.

This is certainly the most cursed inequality regarding quadratics I have ever come across. It seems like the inf of the difference between these function is a smooth (?) function with compact support! Very strange.

Can anyone prove this inequality? The inequality is true for large $x_i$, so the minimum of the difference of RHS-LHS occurs when the gradient is $0$, so one could calculate the gradient and get $n$ polynomial equations and try to find the zeroes, but this does not seem to be a workable solution.

Answer 1

Let $A = \sqrt{a_1^2 + \cdots + a_n^2}$ and $B = \sqrt{x_1^2 + \cdots + x_n^2}$.

Using Cauchy-Bunyakovsky-Schwarz inequality, we have $$ (a_1x_1 + \cdots + a_n x_n)^2 \le (a_1^2 + \cdots + a_n^2)(x_1^2 + \cdots + x_n^2) $$ which results in $$a_1x_1 + \cdots + a_n x_n \ge - AB.$$

We have \begin{align*} (x_1 + a_1)^2 + \cdots + (x_n + a_n)^2 = A^2 + B^2 + 2a_1x_1 + \cdots + 2a_n x_n \ge A^2 + B^2 - 2AB. \end{align*}

It suffices to prove that $$A^2 \le (1 + A^2 + B^2 - 2AB)(1 + B^2)$$ or $$(AB - B^2 - 1)^2 \ge 0.$$

We are done.

Answer 2

For $n = 1$, you can see that: $$(1+x_1(x_1+a_1))^2\geq 0$$ gives your inequality. For the inductive hypothesis, assume the inequality for $n-1.$ If one of $a_i$ is zero, say $a_n = 0, $ then: $$a_1^2 + \dots + a_n^2 =a_1^2 + \dots + a_{n-1}^2\leq \left(1+\sum_{i=1}^{n-1} x_i^2\right)\left(1+\sum_{i=1}^{n-1} (x_i+a_i)^2\right)\leq $$ $$\left(1+\sum_{i=1}^{n} x_i^2\right)\left(1+\sum_{i=1}^{n} (x_i+a_i)^2\right)$$ since $x_n + a_n = 0$ and $x_n^2\geq 0.$ So the harder case is when no $a_i$ is zero. In that case, fix $a$ and we want to minimize the RHS over $\mathbb{R}^n$ and its critical points are found from the equations: $$(1+x_1^2 + x_2^2 + \ldots + x_n^2)(x_i + a_i) + (1+(x_1+a_1)^2 + \ldots + (x_n+a_n)^2)x_i = 0$$ for all $1\leq i\leq n.$ This implies: $$\dfrac{x_1+a_1}{x_1} = \dfrac{x_2+a_2}{x_2}= \dots \dfrac{x_n+a_n}{x_n} = c\neq 0.$$ If $c = 0$, $x_1 + a_1 = 0$ for example and this would mean $x_1 = 0$, which implies $a_1 = 0$ and it contradicts our assumption. Therefore, $$x_i(1 +c)(1 + \sum x_i^2) = 0\implies c = -1\implies x_i = -\dfrac{a_i}{2}.$$ For these values, the inequality to be proved is: $$\sum a_i^2\leq (1 + \dfrac 14\sum a_i^2)^2\iff 4A \leq (1 + A)^2\iff (A-1)^2\geq 0 $$ and we are done.