Let $a,b \geq 0$, and $n \in\mathbb{N}$, then if $1<p,q <\infty$ with $\frac{1}{p}+\frac{1}{q}=1$ is valid
$(a+b)^n \leq p^{n}a^{n}+q^{n}b^{n}$.
We would like to know in place of exponent $n$ there was a variable exponent $f(x)$ with $f:B^{1}\to \mathbb{R}$ function. What conditions should the function p have to have a type 1 inequality
$(a+b)^{f(x)} \leq p^{n}a^{f(x)}+q^{n}b^{f(x)}$.?
$B_{1}$ is the unit ball in $\mathbb{R}^{d}$
By Holder $$p^na^n+q^nb^n=\left(p^na^n+q^nb^n\right)\left(\frac{1}{p}+\frac{1}{q}\right)^n\geq\left(a^{\frac{n}{n+1}}+b^{\frac{n}{n+1}}\right)^{n+1}.$$ Id est, it's enough to prove that $$a^{\frac{n}{n+1}}+b^{\frac{n}{n+1}}\geq(a+b)^{\frac{n}{n+1}},$$ which is just Karamata for the concave function $f(x)=x^{\frac{n}{n+1}}.$
Indeed, let $a\geq b$.
Thus, $$(a+b,0)\succ(a,b),$$ which gives $$f(a+b)+f(0)\leq f(a)+f(b).$$