How to prove ${{a}^{a}}{{b}^{b}}\ge {{\left(\frac{a+b}{2}\right)}^{a+b}}$ ?thanks. How to prove 
            $${{a}^{a}}{{b}^{b}}\ge {{\left(\frac{a+b}{2}\right)}^{a+b}}$$
$a>0$,$b>0$,
thanks.

This problem can also be approached by taking weighted AM > GM inequality on $ \left(\frac{a+b}{2a}\right)$ with weight $ a$ and $\left(\frac{a+b}{2b}\right)$ with weight $b$. Then we arrive at $$ \left(\dfrac{a \times \dfrac{a+b}{2a} + b \times \dfrac{a+b}{2b}}{a+b} \right)^{a+b}\ge \left(\frac{a+b}{2a}\right)^{a}\left(\frac{a+b}{2b}\right)^{b}$$ or, $$ 1 \ge \left(\frac{a+b}{2a}\right)^{a}\left(\frac{a+b}{2b}\right)^{b}$$ or, $$ a^a b^b\ge \left(\frac{a+b}{2}\right)^{a+b}$$
