Let $0 \leq a \leq b, 0 \leq k \leq e$ $$ a^{ka}+b^{kb} \geq a^{kb}+b^{ka} $$
It's relatively easy to prove when $b \geq 1$(every non-negative $k$ satisfies this inequality), I can't prove the other case($b \leq 1$).
(Maximum of $k$ is calculated by computer to be about 2.7, so I conjectured it to $e$ but it may be wrong)
Also, generally, let $0 \leq a \leq b, 0 \leq c \leq d$ $$ a^{c}+b^{d} \geq a^{d}+b^{c} $$ When $b \geq 1$, every $c$ and $d$ satisfies this inequality. However, when $b \leq 1$, is there general conditions for $c$ and $d$?
This is equivalent to $$b^{kb}-b^{ka}\ge a^{kb}-a^{ka}$$ Dividing, note that the RHS is negative ($\alpha^{x}$ is decreasing for $\alpha \in (0,1)$), so we want to show $$\frac{b^{kb}-b^{ka}}{a^{kb}-a^{ka}}\le 1$$ Let $f(x)=b^{kx}$ and $g(x)=a^{kx}$. Then by Cauchy's Mean Value Theorem, there is an $c \in (a,b)$ with $$\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(c)}{g'(c)}=\frac{\log(b)}{\log(a)}\left(\frac{b}{a}\right)^{kc}$$ We want to prove this is at most $1$, for some restriction on $k$. Since $b \ge a$, $(b/a)^{kc}$ is maximised at $c=b$. So consider $h(x)=\log(x) x^{kb}$. When $h$ is increasing,
$$\frac{f(b)-f(a)}{g(b)-g(a)}\le \frac{h(b)}{h(a)}\le 1$$
$h$ is increasing when $h'(x)>0$. By computation, $h'(x)=(1-kb)x^{kb-1}$, which is positive for positive $x$ iff $1-kb>0$, i.e. $k<1/b$. I think these are the best bounds you will get.