# An inequality in an Euclidean space

For $$n\geq 1$$, endow $$\mathbb{R}^n$$ with the usual scalar product. Let $$u=(1,1,\dots,1)\in\mathbb{R}^n$$, $$v\in {]0,+\infty[^n}$$ and denote by $$p_{u^\perp}$$ and $$p_{v^\perp}$$ the orthographic projection onto the hyperplanes $$u^\perp$$ and $$v^\perp$$. For $$x\in\mathbb{R}^n$$ such that $$x\notin [0,+\infty[^n$$ and $$x\notin {]-\infty,0]^n}$$, do the following hold:

$$p_{u^\perp}(x)\cdot p_{v^\perp}(x)>0?$$

• I am never sure what symbols such as $\Bbb R_+$ and $\Bbb R_+^*$ mean. Apr 19 at 19:19
• I edited, thank you. Apr 19 at 19:29

Imagine that $$v$$ is (almost) equal to $$(1,0,\ldots,0)$$, then for $$x=(x_1,\ldots,x_n)$$, $$\sum x_i=:na$$, we should check that $$(x_1-a,x_2-a,\ldots,x_n-a)\cdot(0,x_2,\ldots,x_n)\geqslant 0$$, in other words $$x_2^2+\ldots+x_n^2-a(x_2+\ldots+x_n)\geqslant 0\,\,$$ (A).

Assume that $$n\geqslant 3$$, fix $$x_2,\ldots,x_n$$ so that there are both positive and negative numbers between them and $$x_2+\ldots+x_n>0$$. Then for very large $$x_1$$ the guy $$a$$ is very large too, and (A) does not hold

It doesn't hold, although it looks quite rare. For example $$u:=\begin{pmatrix}1\\1\\1\end{pmatrix}, \quad v:=\begin{pmatrix}\;1\\\;1\\27\end{pmatrix}, \quad x:=\begin{pmatrix}-1\\\;\;6\\\;25\end{pmatrix}$$

\begin{align} \\ \alpha &:= \frac{x\cdot u}{u\cdot u} &&= \frac{-1\cdot 1+6\cdot 1+25\cdot 1}{1^2+1^2+1^2} &&= \frac{30}{3} &&= 10\\ \\ \beta &:= \frac{x\cdot v}{v\cdot v} &&= \frac{-1\cdot 1+6\cdot 1+25\cdot 27}{1^2+1^2+27^2} &&= \frac{680}{731} &&= \frac{40}{43} \end{align} $$p_{u^\perp}(x) = x-\alpha u = \begin{pmatrix}-11\\-4\\\;15\end{pmatrix}, \qquad p_{v^\perp}(x) = x-\beta v = \begin{pmatrix}-\frac{83}{43}\\ \;\;\frac{218}{43}\\ -\frac{5}{43}\end{pmatrix}$$

$$p_{u^\perp}(x)\cdot p_{v^\perp}(x) \quad=11\cdot\frac{83}{43} - 4\cdot \frac{218}{43} - 15\cdot\frac{5}{43} \quad=-\frac{34}{43} \quad<0$$

There is nothing special about $$\mathbf{u}=(1,\dots,1)$$. I work with an arbitrary vector in $$(0,\infty)^n$$.

Claim) Let $$n\geq 3$$ and $$\mathbf{u}=(u_1,\dots,u_n),\mathbf{v}=(v_1,\dots,v_n)\in (0,\infty)^n$$ be two linearly independent vectors. Suppose $$\mathbf{u},\mathbf{v}$$ satisfy the followings:

• $$\frac{u_1}{u_2}<\frac{v_1}{v_2}$$, $$\frac{u_2}{u_3}>\frac{v_2}{v_3}$$;
• $$\frac{A}{B}>\frac{4u_3v_3}{|u_1v_2-u_2v_1|^2}$$ where $$\begin{split} & A:=\langle\mathbf{u},\mathbf{v}\rangle \left(1-\frac{\langle\mathbf{u},\mathbf{v}\rangle^2}{|\mathbf{u}|^2|\mathbf{v}|^2}\right)>0,\\ &B:=|(u_1,u_2,u_3)\times(v_1,v_2,v_3)|^2>0. \end{split} \tag{\star}$$ Then there exists a vector $$\mathbf{x}\in\Bbb{R}^n$$ with both positive and negative components for which $$\left\langle p_{\mathbf{u}^{\perp}}(\mathbf{x}),p_{\mathbf{v}^{\perp}}(\mathbf{x})\right\rangle<0$$.

To see this, consider the quadratic form $$\begin{split} &Q(\mathbf{x}):=\left\langle p_{\mathbf{u}^{\perp}}(\mathbf{x}),p_{\mathbf{v}^{\perp}}(\mathbf{x})\right\rangle= \left\langle\mathbf{x}-\frac{\langle\mathbf{x},\mathbf{u}\rangle}{|\mathbf{u}|^2}\mathbf{u}, \mathbf{x}-\frac{\langle\mathbf{x},\mathbf{v}\rangle}{|\mathbf{v}|^2}\mathbf{v}\right\rangle\\ &=|\mathbf{x}|^2-\frac{|\langle\mathbf{x},\mathbf{u}\rangle|^2}{|\mathbf{u}|^2}-\frac{|\langle\mathbf{x},\mathbf{v}\rangle|^2}{|\mathbf{v}|^2}+\frac{\langle\mathbf{x},\mathbf{u}\rangle\langle\mathbf{x},\mathbf{v}\rangle}{|\mathbf{u}|^2|\mathbf{v}|^2}\langle\mathbf{u},\mathbf{v}\rangle. \end{split}$$ on $$\Bbb{R}^n$$. If $$\mathbf{x}$$ is a linear combination of the form $$\mathbf{u}+t\mathbf{v}$$, then $$Q(\mathbf{u}+t\mathbf{v})= \left\langle t\left(\mathbf{v}-\frac{\langle\mathbf{u},\mathbf{v}\rangle}{|\mathbf{u}|^2}\mathbf{u}\right), \mathbf{u}-\frac{\langle\mathbf{u},\mathbf{v}\rangle}{|\mathbf{v}|^2}\mathbf{v}\right\rangle =t\overbrace{\langle\mathbf{u},\mathbf{v}\rangle}^{>0} \overbrace{\left(-1+\frac{\langle\mathbf{u},\mathbf{v}\rangle^2}{|\mathbf{u}|^2|\mathbf{v}|^2}\right)}^{<0}.\tag{\star\star}$$ When $$n=2$$, the vectors $$\mathbf{u},\mathbf{v}$$ are in the first quadrant. For $$t>0$$, $$Q$$ is negative at $$\pm(\mathbf{u}+t\mathbf{v})$$ which belong to the first or the third quadrant. On the other hand, in order for $$\pm(\mathbf{u}+t\mathbf{v})$$ to belong to either the second or fourth quadrants, $$t$$ should be negative. But then $$Q$$ becomes positive by $$(\star\star)$$. This shows that when $$n=2$$ we have
$$\left\langle p_{\mathbf{u}^{\perp}}(\mathbf{x}),p_{\mathbf{v}^{\perp}}(\mathbf{x})\right\rangle\geq 0$$ if $$\mathbf{x}\notin [0,\infty)^2\cup (-\infty,0]^2$$.

Finally, suppose $$n>2$$. Then there exist non-zero vectors perpendicular to both $$\mathbf{u}$$ and $$\mathbf{w}$$. For any such vector $$\mathbf{w}$$, the formula for $$Q$$ implies $$Q(\mathbf{x}+\mathbf{w})=Q(\mathbf{x})+|\mathbf{w}|^2$$. The idea is to find $$t>0$$ and $$\mathbf{w}\perp\mathbf{u},\mathbf{v}$$ such that $$\mathbf{u}+t\mathbf{v}+\mathbf{w}$$ has both positive and negative components (which should be possible since components of $$\mathbf{w}$$ cannot be all positive or all negative) and
$$Q(\mathbf{u}+t\mathbf{v}+\mathbf{w}) =\overbrace{t\langle\mathbf{u},\mathbf{v}\rangle}^{>0} \overbrace{\left(-1+\frac{\langle\mathbf{u},\mathbf{v}\rangle^2}{|\mathbf{u}|^2|\mathbf{v}|^2}\right)}^{<0}+|\mathbf{w}|^2<0. \tag{\star\star\star}$$ We take $$\mathbf{w}$$ to be $$\mathbf{w}:=s(u_2v_3-u_3v_2,u_3v_1-u_1v_3,u_1v_2-u_2v_1,\overbrace{0,\dots,0}^{n-3})\quad s>0.$$ Clearly, $$\mathbf{w}\perp\mathbf{u},\mathbf{v}$$, and by our hypotheses, its third component is negative while the first one is positive. For any $$s,t>0$$ the first component of $$\mathbf{u}+t\mathbf{v}+\mathbf{w}$$ is positive, and it suffices to arrange $$s,t$$ so that the third component is negative, and moreover $$(\star\star\star)$$ holds. The former amounts to $$u_3+tv_3 while, with the notation from $$(\star)$$, the latter means $$-tA+s^2B<0.$$ So the problem boils down to $$\frac{u_3+tv_3}{|u_1v_2-u_2v_1|}<\sqrt{\frac{tA}{B}}$$, because then any $$s$$ between them qualifies. The last inequality means that
$$r\mapsto{\frac{v_3}{|u_1v_2-u_2v_1|}}r^2-\sqrt{\frac{A}{B}}r+\frac{u_3}{|u_1v_2-u_2v_1|}$$ attains negative values over $$(0,\infty)$$. Since all the coefficients of this quadratic are positive except the linear one, this is equivalent to having positive discriminant: $$\frac{A}{B}>\frac{4u_3v_3}{|u_1v_2-u_2v_1|^2};$$ an inequality which was assumed. $$\blacksquare$$

Corollary) Suppose $$n\geq 3$$ and $$\mathbf{u}\in (0,\infty)^n$$. Then there exist vectors $$\mathbf{v}\in (0,\infty)^n$$ and $$\mathbf{x}\in\Bbb{R}^n\setminus([0,\infty)^n\cup(-\infty,0]^n)$$ such that $$\left\langle p_{\mathbf{u}^{\perp}}(\mathbf{x}),p_{\mathbf{v}^{\perp}}(\mathbf{x})\right\rangle<0$$.

To show this, one only needs to pick $$\mathbf{v}=(v_1,\dots,v_n)$$ with positive entries such that $$\frac{u_1}{u_2}<\frac{v_1}{v_2}$$, $$\frac{u_2}{u_3}>\frac{v_2}{v_3}$$ and $$\frac{A}{B}>\frac{4u_3v_3}{|u_1v_2-u_2v_1|^2}$$. If $$v_1>0$$ is constant and $$v_2,v_3$$ get small so that $$\frac{v_2}{v_3}$$ becomes small too, then $$\frac{u_1}{u_2}<\frac{v_1}{v_2}$$, $$\frac{u_2}{u_3}>\frac{v_2}{v_3}$$ hold; and $$\frac{4u_3v_3}{|u_1v_2-u_2v_1|^2}\to 0$$. But quantities $$A$$ and $$B$$ remain bounded from below. Consequently, when $$v_2,v_3\to 0^{+}$$ such that $$\frac{v_2}{v_3}<\frac{u_2}{u_3}$$ as $$v_1,v_4,\dots,v_n$$ are constant, we obtain vectors $$\mathbf{v}$$ satisfying the desired properties. $$\blacksquare$$

This argument also provides a recipe for constructing vectors $$\mathbf{x}$$ and $$\mathbf{v}$$ for which the dot product under consideration is negative.

• This doesn't seem to work (the projections of $x$ to $u^\perp$ and $v^\perp$ are both in the second quadrant). In general, it seems the statement is true for $n=2$ Apr 19 at 19:23
• Sorry, I had the projection along $u$ and $v$ in mind. Apr 19 at 19:25
• @SaúlRM I rewrote my answer. Apr 20 at 1:43