Solution:

We start with a general fact about any positive real number $x$. There are two cases: either $x \geq 1$ or $x < 1$.

- If $x \geq 1$ then $x^{p} \geq x^{q}$ for any $p > q$. Substituting $p = x$ and $q = 1$ gives us $x^{x} \geq x^{1} = x$.

- If $x < 1$ then $x^{p} < x^{q}$ for any $p > q$. Substituting $p = 1$ and $q = x$ gives us $x = x^{1} < x^{x}$.

In either case we get $x^{x} \geq x$ for all positive real numbers $x$. Applying this fact for $a$, $b$ and $c$ gives us
$$a^{a} + b^{b} + c^{c} \geq a + b + c = 3$$
as required.