I know that $x,y\in \mathbb{R}^n$ are such that $x_1\leq0,x_1^2\geq x_2^2+\dots+x_n^2$ and $y_1\geq 0,y_1^2\geq y_2^2+\dots+y_n^2$. Is it possible to show that $$x_1y_1+x_2y_2+\dots +x_ny_n\leq 0$$

Clearly, $x_1y_1\leq 0$, but my all attempts to show that $x_2y_2+\dots +x_ny_n\leq 0$ failed.


2 Answers 2


Set $x^\prime \stackrel{\rm{}def}{=}(x_2,\dots,x_n)\in\mathbb{R}^{n-1}$, $y^\prime \stackrel{\rm{}def}{=}(y_2,\dots,y_n)\in\mathbb{R}^{n-1}$. You have \begin{align*} x_2^2+\dots x_n^2 = \lVert x^\prime \rVert^2 &\leq x_1^2 \\ y_2^2+\dots y_n^2 = \lVert y^\prime \rVert^2 &\leq y_1^2 \\ \end{align*} and by Cauchy-Schwarz $$ x_2 y_2+\dots x_n y_n = \langle x^\prime,y^\prime\rangle \leq \lVert x^\prime \rVert\lVert y^\prime \rVert \leq \vert{x_1}\rvert\lvert{y_1}\rvert = -x_1 y_1 $$ therefore $$ x_1 y_1 + x_2 y_2+\dots x_n y_n \leq 0.$$


Following from the proof given by @ClementC. one can strengthen the bounds using a similar approach as proving the Cauchy Schwarz inequality,

Consider, $f(x) = (y_1^2-y_2^2-\cdots -y_n^2)x^2-2(x_1y_1-x_2y_2-\cdots-x_ny_n)x+(x_1^2-x_2^2-\cdots -x_n^2)=(y_1x-x_1)^2-(y_2x-x_1)^2-\cdots -(y_nx-x_n)^2$. Now taking $x=\frac{x_1}{y_1}$, we get $f(\frac{x_1}{y_1})=-(y_2\frac{x_1}{y_1}-x_1)^2-\cdots -(y_n\frac{x_1}{y_1}-x_n)^2\le 0$. However the leading expression of the quadratic $f$ is positive. Therefore, $f(x)\rightarrow \infty$ as $x\rightarrow \pm\infty$. Since, $f(x)\le 0$, the equation $f(x)=0$, has one root each in the intervals $(-\infty,\frac{x_1}{y_1}]$ and $[\frac{x_1}{y_1},-\infty)$. Hence the discriminant of $f$, must be non-negative, giving $0\le (y_1^2-y_2^2-\cdots -y_n^2)(x_1^2-x_2^2-\cdots -x_n^2)\le (x_1y_1-x_2y_2-\cdots -x_ny_n)^2$.

That is, $|x_1y_1+x_2y_2+\cdots +x_ny_n-2x_1y_1|\ge \sqrt{(y_1^2-y_2^2-\cdots -y_n^2)(x_1^2-x_2^2-\cdots -x_n^2)}$

or, $x_1y_1+x_2y_2+\cdots +x_ny_n \le 2x_1y_1 - \sqrt{(y_1^2-y_2^2-\cdots -y_n^2)(x_1^2-x_2^2-\cdots -x_n^2)} \le 0$.


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