I used in the proof of Hlawka's Inequality you can find the link here Hlawka's Inequality
that's if i have function of three variable is even in each variable, so that :
$$f(a,b,c)=f(|a|,|b|,|c|)$$
but a teacher that I know told me why you generalize it for the three variable, I told him that we note $x,y$ and $z$ play symmetric roles in the inequality he said even that you don't have right to say $$f(a,b,c)=f(|a|,|b|,|c|)$$
I need you to convince him by a theorem or by reference in a book I don't know what to do
any help would be appreciated !
When a function $$g:\quad {\mathbb R}\to{\mathbb R},\qquad t\mapsto g(t)$$ is even then for $t\geq0$ one has $g(t)=g\bigl(|t|\bigr)$, and for $t<0$ one has $g(t)=g(-t)=g\bigl(|t|\bigr)$. Therefore $$g(t)=g\bigl(|t|\bigr)\qquad\forall t\in{\mathbb R}\ .$$ When $f:\>{\mathbb R}^3\to{\mathbb R}$ is even in each variable then, according to this principle, for fixed $b$, $c$ the function $$\phi(a):=f(a,b,c)$$ satisfies $$\phi(a)=\phi\bigl(|a|\bigr)\qquad\forall a\in{\mathbb R}\ .$$ It follows that for all $b$, $c$ we have $$f(a,b,c)=f\bigl(|a|,b,c\bigr)\ .\tag{1}$$ Now, for fixed $a$, $c$ the left side of $(1)$ by assumption, and therefore also the right side of $(1)$, are even functions of $b$. It follows that $$f\bigl(|a|,b,c\bigr)=f\bigl(|a|,|b|,c\bigr)\qquad \forall b\in{\mathbb R}\ ,$$ and on and on. All this is a complicated way of saying that you were right in the first place.