An algebra problem by Ace star

Observe that $f(x) = x^x$ is convex over all $\mathbb R^+$ . Then, due to Jensen's Inequality, using $f(x) = x^x$ :

$\begin{aligned} \dfrac{x^x+y^y}{2} &\ge \left(\dfrac{x+y}{2}\right)^{\frac{x+y}{2}}\\ x^x + y^y &\ge 2\left(\dfrac{1}{2}\right)^{\frac{1}{2}}\\ x^x + y^y &\ge \boxed{\sqrt{2}} \end{aligned}$

By the AM-GM inequality , $x^x + y^2 \geq 2\sqrt{x^x y^y}$ with a minimum when $x^x = y^y$ , or in other words, when $x = y$ .

Since the sum of $x$ and $y$ is $1$ , $x = y = \frac{1}{2}$ , and the minimum value is $\frac{1}{2}^{\frac{1}{2}} + \frac{1}{2}^{\frac{1}{2}} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \approx \boxed{1.414}$ .

Lazy Lagrange Multipliers at work here! Let $f(x,y) = x^x + y^y$ and $g(x,y) = x+y=1$ such that $\nabla f = \lambda \nabla g$ , or:

$x^x (\ln(x)+1) = \lambda;$

$y^y(\ln(y)+1) = \lambda$

which requires $y = x$ . Thus, $x + x = 1 \Rightarrow x = y = \frac{1}{2}.$ The Hessian matrix of $f(x,y)$ computes to:

$F(x,y) = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix} = \begin{bmatrix} x^{x-1} + x^x[\ln^{2}(x)+1] & 0 \\ 0 & y^{y-1} + y^y[\ln^{2}(y)+1] \end{bmatrix}$ ;

or $F(1/2,1/2) = [\sqrt{2} + \frac{1}{\sqrt{2}} \cdot [1+\ln^{2}(1/2)]] \cdot I_{2x2} > 0$ , which is positive-definite and yields a global minimum at $(x,y) = (1/2,1/2).$ The minimum value is finally determined to be $f(1/2,1/2) = 2(1/\sqrt{2}) = \boxed{\sqrt{2}}.$

$y=x^x+(1-x)^{1-x}$ is symmetrical over $x=\frac12$ . Because $x^x$ is convex, the symmetry entails $y$ is minimum at $x=\frac12$ . $\left(\frac12\right)^{\frac12}+\left(\frac12\right)^{\frac12}=\sqrt2$ .

Note that .5+.5=1 is the only choice where the values are not symmetrical.

For example, .1+.9=1 and .9+.1=1 so

.5^.5 + .5^.5 = 1.414

.6^.6 + .4^.4 = 1.429

.8^.8 + .2^.2 = 1.561

Or graph x^x + (1-x)^(1-x) to see that smallest value occurs at 0.5