I'm trying to show this fact, for a generic function $f:E\subset\mathbb{R}\to\mathbb{R}$:
$$ \omega(\delta _1+\delta_2)\leq \omega(\delta _1)+\omega(\delta _2) $$
where $$\omega(\delta):=\sup_{x_1,x_2\in E\\ |x_1-x_2|<\delta}|f(x_1)-f(x_2)|.$$
I have done this:
$$\omega(\delta _1)+\omega(\delta _2) =\sup_{x_1,x_0\in E \\ |x_1-x_0|<\delta_1}|f(x_1)-f(x_0)|+\sup_{x_2,x_0'\in E \\ |x_2-x_0'|<\delta_2}|f(x_0')-f(x_2)| =$$
$$=\sup_{x_1,x_0,x_0',x_2\in E \\ |x_1-x_0|<\delta_1\\ |x_2-x_0'|<\delta_2}|f(x_1)-f(x_0)|+|f(x_0')-f(x_2)| \geq \sup_{x_1,x_0,x_2\in E \\ |x_1-x_0|<\delta_1\\ |x_2-x_0|<\delta_2}|f(x_1)-f(x_0)|+|f(x_0)-f(x_2)| \geq$$
$$\geq \sup_{x_1,x_0,x_2\in E \\ |x_1-x_0|<\delta_1\\ |x_2-x_0|<\delta_2}|f(x_1)-f(x_2)| $$
but from this I can't pass to $\sup_{x_1,x_2\in E \\ |x_1-x_2|<\delta_1+\delta_2}|f(x_1)-f(x_2)| =:\omega(\delta_1+\delta_2) $ because this would introduce a $\leq$ in the inequality chain, and not a $\geq$ (that is needed to conclude).
How can I fix this?
Thanks in advance.