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

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
$$a^{|b-a|}+b^{|c-b|}+c^{|a-c|} &gt; \frac52$$
$$\Leftrightarrow \frac{a^a}{a^b} +\frac{b^b}{b^c}+\frac{c^a}{c^c}&gt; \frac52$$
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 $\,|f(a,b)| &lt; 0.02$ . There seem to be several of these minimum values. I wish the rigorous proof producers among us good luck.
Perhaps it would help to take $a=1+x$, $b=1+y$, $c=1+z$, so that $x+y+z=0$

Now the inequality looks like:

$$(1+x)^{|y-x|} + (1+y)^{|z-y|} + (1+z)^{|x-z|} \geq \frac{5}{2}$$

Using bernoulli's inequality, the expression is more or less

$$1 + x|y-x| + 1 + y|z-y| + 1 + z|x-z| \geq \frac{5}{2}$$
$$x|y-x| + y|z-y| + z|x-z| \geq \frac{-1}{2}$$

And we can split it in two cases depending on how x, y and z are sorted.

For example, if $x&lt;y&lt;z$, we must prove that:
$$x(y-x) + y(z-y) + z(z - x) = -x^2 - y^2 + z^2 + xy + yz - xz \geq \frac{-1}{2}$$

We recall that $x+y+z=0$, and thus $z = -x-y$:

$$-x^2 - y^2 + z^2 + xy + yz - xz = -x^2 - y^2 + x^2 + y^2 - 2xy + xy - (y+x)(y-x) = x^2 - y^2 - xy$$

It should not be hard to check if this is more that $\frac{-1}{2}$ given our restraints.