If $q$ is in $[1,2)$ and $a,b$ real numbers, is it always true that
$|a+b|^q + |a-b|^q \leq 2(|a|^q+|b|^q)$ ?
I can prove $|a+b|^q \leq 2^q (|a|^q + |b|^q)$ but this leaves me with an extra $2^q$ and I am not really sure how to better estimate the second term: $|a-b|^q$ or use the bound on $q$.
If either $a$ or $b$ is zero, the inequality is obviously an equality. So we need to consider only cases where both are non-zero. WLOG, let $a \ge b > 0$ so that $x = \frac{b}{a} \in [0, 1]$. Then it is sufficient to show:
$$(1+x)^q + (1-x)^q \leq 2(1+x^q)$$
From power means, we have for $q \le 2$: $$\sqrt[q]{\frac{(1+x)^q+(1-x)^q}{2}} \le \sqrt{\frac{(1+x)^2+(1-x)^2}{2}} = \sqrt{1+x^2}$$
$$\implies (1+x)^q+(1-x)^q \le 2(1+x^2)^{q/2}$$
So what remains is to show $(1+x^2)^{q/2} \le 1+x^q$ or equivalently to show that $f(r) = \sqrt[r]{1+x^r}$ is a decreasing function, which follows from the sign of $f'( r)$.