<s>[INST] 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.
 [/INST] Let $P=(1+x)^{1+x} (1-x)^{1-x}$ , where $0 ≤ x <1$ ; then 
$\log P=(1+x)\log(1+x)+(1-x)\log(1-x)$
$\space$$\space$ $\space$$=x${$\log(1+x)-\log(1-x)$ } +$\log(1+x)+\log(1-x)$ $\space$$=2x\bigg(x+\frac{x^3}{3} + \frac{x^5}{5}+...\bigg) - 2\bigg(\frac{x^2}{2}+\frac{x^4}{4} + \frac{x^6}{6}+...\bigg)$ $=2\bigg(\frac{x^2}{1×2}+\frac{x^4}{3×4} + \frac{x^6}{5×6}+...\bigg)$$≥0$ 
Hence $\log P≥0$ i.e. $P ≥1$ that is $(1+x)^{1+x} (1-x)^{1-x}≥1$ ; put $x=z/u$ , where $u>z$ ; then 
$\bigg(1+z/u\bigg)^{1+z/u} \bigg(1-z/u\bigg)^{1-z/u}\space≥1$ $\space$ i.e. $\bigg(\frac{u+z}u\bigg)^{u+z} \bigg(\frac{u-z}u\bigg)^{u-z}≥1^u=1$  
i.e. $\big(u+z\big)^{u+z} \big(u-z\big)^{u-z}≥u^{2u}$ ; now put $u+z=a , u-z=b$ , so that 
$u=\frac{a+b}2 , a>0,b>0$ ; 
then $a^ab^b ≥ \bigg(\frac{a+b}2\bigg)^{a+b}$
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