2019 NZMO2 P3

Let f(x) = x^x. We can see that the graph of x^x is always positive having a minima at x =1/e. Basically it is a convex graph. Using Jensen's inequality for a convex graph , we have (f(a) + f(b) + f(c))/3 >= f( (a+b+c)/3). Equality holds when a=b=c. Thus a^a + b^b + c^c} >= 3f(1) as a+b+c=3. f(1)=1^1=1. Hence required answer is 3.

For any positive real number $x$ , there are two possible cases -

$\begin{cases} \text{If } x \geq 1, \text{ then } x^m \geq x^n \text{ for any positive reals }m > n. \text{ So } x^x \geq x, \text{ when }m=x, n=1. \\ \text{If } x < 1, \text{ then } x^m < x^n \text{ for any positive reals }m > n. \text{ So } x < x^x, \text{ when }m=1, n=x. \end{cases}$

In both cases we get $x^x ≥ x$ for all positive real numbers $x$ .

$\therefore a^a+b^b+c^c \geq a+b+c=3 \\ \boxed{\implies a^a+b^b+c^c \geq 3}$

See this .