Sketch: Since $x^x\ge x$: $$a^a+b^b+c^c\ge a+b+c=3.$$

jasperE3 wrote: Sketch: Since $x^x\ge x$: $$a^a+b^b+c^c\ge a+b+c=3.$$ In addition to the above solution, we can also use the inequality Bernoulli. We assume $a\geq b \geq c$. + $a \geq b \geq 1 \geq c$. We have: $${a^a} \ge 1 + a\left( {a - 1} \right) = {a^2} - a + 1;{b^b} \ge 1 + b\left( {b - 1} \right) = {b^2} - b + 1,$$and $${\left( {\frac{1}{c}} \right)^c} \le 1 + c\left( {\frac{1}{c} - 1} \right) = 1 + 1 - c = 2 - c \Rightarrow {c^c} \ge \frac{1}{{2 - c}}.$$We need: $${a^2} + {b^2} - \left( {a + b} \right) + 2 + \frac{1}{{2 - c}} \ge 3.$$According to the AM - GM inequality, we have: $$ \begin{aligned} a^{2}+b^{2}-(a+b)+2+\frac{1}{2-c} & \geq \frac{(a+b)^{2}}{2}-(3-c)+2+\frac{1}{2-c} \\\\ &=\frac{(3-c)^{2}}{2}-(3-c)+2+\frac{1}{2-c}. \end{aligned} $$So we only need to prove : \[\frac{{{{\left( {3 - c} \right)}^2}}}{2} - \left( {3 - c} \right) + 2 + \frac{1}{{2 - c}} \ge 3 \Leftrightarrow \frac{{(4 - c){{(c - 1)}^2}}}{{2(2 - c)}} \ge 0.\]+ $a \geq 1 \geq b \geq c$. Same as above, we have: \[{a^a} \ge {a^2} - a + 1,{b^b} \ge \frac{1}{{2 - b}},{c^c} \ge \frac{1}{{2 - c}}.\]We need: \[{a^2} - a + 1 + \frac{1}{{2 - b}} + \frac{1}{{2 - c}} \ge 3.\]According to the Cauchy – Schwarz inequality \[{a^2} - a + 1 + \frac{1}{{2 - b}} + \frac{1}{{2 - c}} \ge {a^2} - a + 1 + \frac{4}{{4 - b - c}} = {a^2} - a + 1 + \frac{4}{{1 + a}}.\]So we only need to prove : \[{a^2} - a + 1 + \frac{4}{{1 + a}} \ge 3 \Leftrightarrow \frac{{{{(a - 1)}^2}(a + 2)}}{{a + 1}} \ge 0.\]equality occurs if and only if $a=b=c=1$.

$$\text{ Let }f(x)=x^x\text{ now notice that if we take the first and second derivative we obtain }$$$$f'(x)=x^x\ln x+x^x\text{ and }f''(x)=x^x+x^{x-1}+x^x\ln^2x+2x^x\ln x>0, \forall x\in\mathbb{R}^+ \text{ therefore }f \text{ is convex }$$$$\text{ Furthermore }\sum_{cyc}a^a=\sum_{cyc}f(a)\overset{\text{Jensen}}{\ge}3f\left(\frac{\sum_{cyc}a}{3}\right)=3f(1)=3\cdot1^1=3$$$$\blacksquare.$$