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.
 A: 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.
A: 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 $a_2=\dfrac{1}{b}$ , $q_2=b$
$$\dfrac{1}{a^ab^b} \le (\dfrac{2}{a+b})^{a+b}$$
$$a^ab^b \ge (\dfrac{a+b}{2})^{a+b}$$
A: You can use the weighted AM $\ge$ GM inequality.
$$\left(\dfrac{a \times \dfrac{1}{a} + b \times \frac{1}{b}}{a+b} \right)^{a+b}\ge \dfrac{1}{a^a b^b}$$
$$\left(\dfrac{2}{a+b} \right)^{a+b}\ge \dfrac{1}{a^a b^b}$$
A: Since $\ln$ is an increasing function, so it is enough to show that, after taking log of both sides:
$$\frac{a\ln a+b\ln b}{a+b}\geq \ln\left(\frac{a+b}{2}\right).$$
Now consider the graph of $y=x\ln x.$ Note that $y''=(1+\ln x)'=\frac 1x>0$ for all $x>0,$ and therefore it is convex. So applying Jensen's inequality we have
$$a\ln a +b\ln b\geq 2\left(\frac{a+b}{2}\right)\ln\left(\frac{a+b}{2}\right),$$
As desired. Equality holds iff $a=b.\Box$
A: 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 $a^ab^b ≥ \bigg(\frac{a+b}2\bigg)^{a+b}$
A: Use GM $\ge$ HM for postive numbers (see here)
A: The function $f(x)=x\log(x)$ is convex for $x\gt0$ since $f''(x)=\frac1x\gt0$. A relation that is often used to define convex functions is
$$
f\!\left(\frac{a+b}2\right)\le\frac{f(a)+f(b)}2\tag1
$$
Applying the definition of $f(x)$ to $(1)$ gives
$$
\left(\frac{a+b}2\right)\log\!\left(\frac{a+b}2\right)\le\frac{a\log(a)+b\log(b)}2\tag2
$$
Each side of $(2)$ is half the log of the corresponding side of
$$
\left(\frac{a+b}2\right)^{a+b}\le a^ab^b\tag3
$$
A: 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, $$ a^a b^b\ge \left(\frac{a+b}{2}\right)^{a+b}$$
