For real numbers $x_1,\ldots,x_n$ and $y_1,\ldots,y_n$ with $$x_1,y_1>0,\ x_1^2>x_2^2+\ldots+x_n^2,\ y_1^2>y_2^2+\ldots+y_n^2,$$ show that $$x_1y_1-x_2y_2-\ldots-x_ny_n \geq \sqrt{(x_1^2-x_2^2-\ldots-x_n^2)(y_1^2-y_2^2-\ldots-y_n^2)}$$ and determine necessary and sufficient conditions for equality to hold.

So this inequality (perhaps obviously) comes from Lorentzian linear algebra. In particular, the Cauchy-Schwarz inequality for the Lorentz norm. The proof I saw actually simplified the problem considerably by assuming without loss of generality that $y=(1,0,0,\ldots,0)$ (if I recall correctly). This 'wlog' was justified by a lemma we had proven earlier, that the positive Lorentz group acts transitively on $m$-dimensional time-like subspaces of $n$-dimensional hyperbolic space, which itself was somewhat involved, using Gram-Schmidt orthonormalisation in the proof.

That said, I would like to know whether there are any direct algebraic proofs of this, since in itself it is just an elementary algebra problem.I would be surprised if the properties of hyperbolic isometries were really necessary just to prove this inequality. I hope I haven't made a mistake in the problem statement.


1 Answer 1


Indeed there is a simple direct proof. Let $\alpha =\sqrt {x_2^{2}+x_3^{2}+\cdots +x_n^{2}}$ and $\beta =\sqrt {y_2^{2}+y_3^{2}+\cdots +y_n^{2}}$. The inequality reduces to $x_1y_1 -\alpha \beta \geq \sqrt {x_1^{2}-\alpha^{2}} \sqrt{y_1^{2}-\beta^{2}}$ which reduces to $2x_1y_1\alpha \beta \leq x_1^{2}\beta ^{2}+y_1^{2}\alpha^{2}$ which is obvious.

  • $\begingroup$ That really is a lot simpler! $\endgroup$
    – AlephNull
    Commented Jan 2, 2019 at 23:37

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