How to prove ${{a}^{a}}{{b}^{b}}\ge {{\left(\frac{a+b}{2}\right)}^{a+b}}$ ?thanks.

Dividing by $b^{a+b}$ and taking the $b^{th}$ root, we need to prove $$\left(\dfrac{a}{b} \right)^{a/b} \geq \left(\dfrac{a/b+1}2 \right)^{a/b+1}$$ Let $a/b = t$. We then need to prove that $$t^t \geq \left(\dfrac{1+t}2\right)^{1+t}$$ Consider the function $$f(x) = x \log(x) - (1+x) \log \left(\dfrac{1+x}2 \right)$$ We then have \begin{align} f'(x) & = x \cdot \dfrac1x + \log(x) - (1+x) \cdot \dfrac1{1+x} - \log(1+x) + \log(2)\\ & = \log(2)+ \log(x) - \log(1+x) = \log\left(\dfrac{2x}{1+x}\right)\\ f''(x) & = \dfrac1x - \dfrac1{1+x} = \dfrac1{x(1+x)} \end{align} Hence, we see that for $x>0$, we have $f(x)$ to attain an extremum when $$f'(x) = 0 \implies 2x = 1+x \implies x = 1$$ And this extremum is a minimum since $f''(x) = \dfrac1{x(1+x)} > 0$ for $x>0$. Hence, we have $$f(x) \geq f(1) = 0$$ Hence, $f(x)$ is non-negative i.e. $f(x) \geq 0$ for all $x \in \mathbb{R}^+$. Now take $x = \dfrac{a}b$ and do some algebraic massaging to get your answer.

Using Power mean inequality

$$\prod_{i=1}^n a_i^{q_i} \le (\dfrac{ \sum_{i=1}^n a_iq_i}{\sum_{i=1}^nq_i})^{\sum_{i=1}^n q_i}$$

Take $a_1=\dfrac{1}{a}$ , $q_1=a$ and $a_2=\dfrac{1}{b}$ , $q_2=b$

$$\dfrac{1}{a^ab^b} \le (\dfrac{2}{a+b})^{a+b}$$

$$a^ab^b \ge (\dfrac{a+b}{2})^{a+b}$$

You can use the weighted AM $\ge$ GM inequality.

$$\left(\dfrac{a \times \dfrac{1}{a} + b \times \frac{1}{b}}{a+b} \right)^{a+b}\ge \dfrac{1}{a^a b^b}$$

$$\left(\dfrac{2}{a+b} \right)^{a+b}\ge \dfrac{1}{a^a b^b}$$