Unexpectedly Powerful

Minimum

Let $f (x,y,z) = x^xy^yz^z$ . Let $g (x) = x \text{ln}x$ .
We can show that $g (x)$ is a convex function by considering its second derivative.
By jensen's inequality ,

$ln (f (x,y,z)) = g (x) + g (y) + g (z) \geq 0$

Hence, $f(x,y,z) \geq e ^ 0 = 1$ .

Maximum

First, we show that $f (x,y,z) \leq f (x+y-0.5, 0.5, z)$ . This is equivalent to proving

$(a+b+\frac {1}{2})^{(a+b+\frac {1}{2})} \geq \sqrt {2} ( (a+\frac {1}{2})^{(a+\frac {1}{2})} (b+\frac {1}{2})^{(b+\frac {1}{2})})$

For nonnegative $a,b$

Now, assume WLOG $a \geq b \geq 0$ . Since $b \geq 0$

$(a+b+\frac {1}{2})^{(a+b+\frac {1}{2})} = (a+b+\frac {1}{2})^{a} (a+b+\frac {1}{2})^{b} (a+b+\frac {1}{2})^{\frac {1}{2}} \geq (a+\frac {1}{2})^{a} (a+b+\frac {1}{2})^{b} (a+b+\frac {1}{2})^{\frac {1}{2}}$

We just then need to prove that

$(a+b+\frac {1}{2})^{(b+\frac {1}{2})} \geq \sqrt {2} ( (a+\frac {1}{2})^{(\frac {1}{2})} (b+\frac {1}{2})^{(b+\frac {1}{2})})$

Or that

$\frac{(1+\frac{a}{b+0.5})^b(1+\frac{b}{a+0.5})^{0.5}}{\sqrt{2b+1}} \geq 1$

Which was attained by dividing the LHS by the RHS. But then, by Bernoulli's inequality, $(1+a)^n \geq 1+na$ , we get

$\frac{(1+\frac{a}{b+0.5})^b(1+\frac{b}{a+0.5})^{0.5}}{\sqrt{2b+1}} \geq \frac{(1+\frac{2ab}{2b+1})(1+\frac{b}{2b+1})}{\sqrt{2b+1}}$

Now we need to prove

$\frac{b(2a+2+\frac{1}{b})(2a+1+y)}{(2a+1)(2b+1)^{1.5}} \geq 1$ .

Since equality occurs when $b=0$ and for all positive b, the LHS is an increasing function in a, we can say that the minimum of the LHS occus when a is zero, and thus we just need to prove that

$\frac{b+1}{\sqrt{2b+1}} \geq 1$

Which is true for all nonnegative b. (One can prove it by Bernoulli's inequality!) Thus the original inequality is proven, and equality occurs when $b= 0$ .

Thus,

$f (x,y,z) \leq f (x+y-0.5, 0.5, z) \leq f(x+y+z -1, 0.5, 0.5) = 2$ . So the maximum is $2$ and the minimum is $1$ which gives our answer to be $1+2 = \boxed{3}$ .

Let $u = x^x y^y z^z$ . We want to find when $u$ attains its minimum and maximum values.

• Check the edge of the domain: $x = y = \tfrac12$ , $z = 2$ . Then $u = {\sqrt(1/2)}^2\cdot 2^2 = 2$ .

• Also on the edge of the domain: $x = \tfrac12$ , $y > \tfrac12$ and $z = 2\tfrac12 - y$ . View $u$ as a function of $y$ , we have $\ln u = \tfrac12\ln \tfrac 12 + y\ln y + z\ln z,$ $d\ln u = (1 + \ln y)\ dy + (1 + \ln z)\ dz;$ since $dz = -dy$ , $d\ln u = (1 + \ln y - 1 - \ln z)\ dy = \left(\ln y - \ln z\right)\ dy = \ln \frac{y}{z}\ dy,$ which is zero if numerator and denominator are equal, i.e. $y = z = 1\tfrac14$ . Then $u = \sqrt{5^5/2^{11}} = \sqrt{3125/2048}$ , which lies between 1 and 2.

• With $x,y > \tfrac12$ and $z = 3 - x - y$ , we view $u$ as a function of $x$ and $y$ : $\ln u = x\ln x + y\ln y + z\ln z,$ $d\ln u = (1 + \ln x)\ dx + (1 + \ln y)\ dy + (1 + \ln z)\ dz;$ with $dz = -dx -dy$ : $d\ln u = (\ln x-\ln z)\ dx + (\ln y - \ln z)\ dy;$ at an extremum, both coefficients of $dx$ and $dy$ are zero, implying $x = y = z = 1$ and $u = 1$ .

Thus we find that the minimum value of $u$ is $A = 1$ and the maximum value of $u$ is $B = 2$ . The total is $\boxed{3}$ .