$a^{|b-a|}+b^{|c-b|}+c^{|a-c|} > \frac52$ for $a,b,c >0$ and $a+b+c=3$

Let $a,b,c >0$ with $a+b+c=3$. Prove that $$a^{|b-a|}+b^{|c-b|}+c^{|a-c|} > \frac52.$$

What I did:

It is cyclic inequality so I assume $c= \min\{ a,b,c \}$.
I consider the first case where $a\ge b\ge c$ then $$a^{|b-a|}+b^{|c-b|}+c^{|a-c|} > \frac52$$ $$\Leftrightarrow \frac{a^a}{a^b} +\frac{b^b}{b^c}+\frac{c^a}{c^c}> \frac52$$ I check function $f(x) =x^x$ to see if it is a strictly monotonic function or not. It turns out that it is a concave up function so I get stuck here.

Not a bounty candidate. Just a pictorial comment.
Make an isoline/contour plot in the $(a,b)$-plane of the function: $$ f(a,b) = a^{|b-a|}+b^{|c-b|}+c^{|a-c|} - \frac52 \quad \mbox{with} \quad c=3-a-b $$ Then this is what we get. The blue spots are where $\,|f(a,b)| < 0.02$ . There seem to be several of these minimum values. I wish the rigorous proof producers among us good luck.