TITLE: Proving $x^x+y^y\ge\sqrt2$ when $x,y\in \mathbb R^+$ and $x+y=1$
QUESTION [10 upvotes]: Assuming  $x,y\in \mathbb R^+$,  $x+y=1$ how to prove $$x^x+y^y\ge\sqrt2$$thanks in advance

REPLY [5 votes]: Without Jensen:
Lets minimize the function on in the positive quadrant.
\begin{align}
f(x)&=x^x+(1-x)^{(1-x)}\\
f^\prime(x)&=(1-x)^{(-x)} (-1+x) (1+\log(1-x))+x^x (1+\log(x))
\end{align}
Equating to zero and solving. (You can also verify that the second derivative is positive)
\begin{align}
x&=0.5\\
f(x)&=(0.5)^{0.5}+(0.5)^{0.5}\\
f(x)&=2\times (0.5)^{0.5}\\
f(x)&=(4\times 0.5)^{0.5}\\
f(x)&=\sqrt{2}\\
\therefore~~~~~~ f(x)&\geq \sqrt{2}\quad \forall x\in \mathbb{R}^+
\end{align}