Show that for all $(\tau, \xi) \in \mathbb R^{n+1}$ we have $|(\tau-ia)^2 - |\xi|^2| \ge a(\tau ^2+|\xi|^2+a^2)^{1/2}$

Question

Show that, for all $(\tau, \xi) \in \mathbb R^{n+1}$, $|(\tau-ia)^2 - |\xi|^2| \ge a(\tau ^2+|\xi|^2+a^2)^{1/2}$

This is the exercise 7.4 in the book by Francois Treves. It is just a fundamental inequality used in other applications.

Well, what I can do is just let $f(\tau, \xi)=L.H.S./R.H.S$. and calculate the expression routinely. I got nothing. I don't even want to use calculus to find the minimum of the function, because it is too complicated. Can anyone share some ideas about this question? Thanks a lot!

Answer 1

This way may be more routine and helps you attack a wider range of inequalities.

We can assume $a\ge0$.

Let $f(a)=(L.H.S.)^2-(R.H.S.)^2$ , and view $\tau, \xi $ as fixed variables. Therefore,

$f'(a)=2a{\tau}^2+2a{\xi}^2 \ge 0$, and $f(0)=0 , \forall (\tau,\xi) \in \mathbb R^{n+1}$

Thus we proved the inequality.

Answer 2

You may write $$ |(\tau-ia)^2 - |\xi|^2| =|\tau-|\xi|-ia | |\tau+|\xi|-ia| = |((\tau-|\xi|)^2+a^2)^{1/2}||((\tau+|\xi|)^2+a^2)^{1/2}| $$ $$ \geq ((|\tau|-|\xi|)^2+a^2)^{1/2}|((\tau+|\xi|)^2+a^2)^{1/2}|\geq|a^2|^{1/2}|((\tau+|\xi|)^2+a^2)^{1/2}|\geq a(\tau^2+|\xi|^2+a^2)^{1/2}. $$