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

Solution 1:

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.

Solution 2:

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}