Let $a,b,c$ be the sides of a triangle in any random order and let $A,B,C$ be the angles opposite to these sides respectively. Let the inradius be $r$ and circumradius be $R$. In this question, it was proved that
$$ \frac{a}{A} + \frac{b}{B} \ge \frac{c}{C} \tag 1 $$
This is similar in form to the classical triangle inequality $a+b \ge c$ which made me wonder if the there exists a general inequality from which both follows. However, obvious suspects such as
$$ \frac{a}{A^x} + \frac{b}{B^x} \ge \frac{c}{C^x} \tag 2 $$ is false in general for $x \ne 0,1$ and I haven't been able to unify the classical triangle inequality with $(1)$ but I observed related inequality which seems to hold for all triangles from which the classical triangle inequality follows as a special case.
Conjecture 1: If $0 \le x \le 2\sqrt{2}/R$ then $$ \frac{a}{A^{rx}} + \frac{b}{B^{rx}} \ge \frac{c}{C^{rx}} $$
Conjecture 2: Let $m = \min(a,b,c)$ be the smallest side of the triangle. If $0 \le x \le 1/R$ then $$ \frac{a}{A^{mx}} + \frac{b}{B^{mx}} \ge \frac{c}{C^{mx}} $$
In both these conjectures, the classical triangle inequality corresponds to the case $x=0$. Can this be proved or disproved?
