Proving $x^x+y^y\ge\sqrt2$ when $x,y\in \mathbb R^+$ and $x+y=1$ Assuming  $x,y\in \mathbb R^+$,  $x+y=1$ how to prove $$x^x+y^y\ge\sqrt2$$thanks in advance 
 A: Let's consider the function
$$f(x)=x \ln x +(1-x)\ln(1-x)$$
where $0<x<1$, and then
$$f'(x)=\ln\frac{x}{1-x}$$
$$f'\left(\frac{1}{2}\right)=0$$
where $x_0=\frac{1}{2}$ is the point where the function reaches its minimum. Then
$$f(x)\ge f\left(\frac{1}{2}\right)$$ 
that finally yields
$$x \ln x +y \ln y \ge \ln\left(\frac{1}{2}\right)\tag1$$ 
Now, let's rewrite the left side inequality, use AM-GM inequality and then use $(1)$ 
$$e^{x \ln x}+e^{y\ln y}\ge 2 {\displaystyle e^{\displaystyle\frac{x \ln x+y \ln y}{2}}}\ge2 {\displaystyle e^{\displaystyle\frac{\ln(1/2)}{2}}}=\sqrt{2}$$
Chris. 
A: 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}
A: HINT Prove that $x^x$ is convex in the interval $[0,1]$. Then use Jensen's inequality.
A: Hint: Apply Jensen's Inequality on $f(x)=x^x$.
