If $b>a>0$ , how to prove that $\left(\frac {\sqrt a +\sqrt b}{2}\right )^2 < \frac 1e \left(\frac {b^b} {a^a}\right)^{\frac 1{b-a}}$? If $b>a>0$ , then how to prove that $\left(\dfrac {\sqrt a +\sqrt b}{2}\right )^2 < \dfrac 1e \left(\dfrac {b^b} {a^a}\right)^{\dfrac 1{b-a}}$ ? 
I tried applying the known standard inequalities but without success . Please help . 
 A: Let $b=ax$, where $x>1$.
Hence, we need to prove that 
$$ea\left(\frac{\sqrt{x}+1}{2}\right)^2<\left(\frac{(ax)^{ax}}{a^a}\right)^{\frac{1}{a(x-1)}}$$ or
$$ea\left(\frac{\sqrt{x}+1}{2}\right)^2<\left(\frac{(ax)^{x}}{a}\right)^{\frac{1}{x-1}}$$ or
$$e\left(\frac{\sqrt{x}+1}{2}\right)^2<\left(x^{x}\right)^{\frac{1}{x-1}}$$ 
or $f(x)>0$, where
$$f(x)=x\ln{x}+1-x+2(x-1)\ln\frac{\sqrt{x}+1}{2},$$
which is obvious after two differentiations. 
A: Taking $\ln$ ( natural logarithm) on both sides of the inequality, we need to prove that $2 \ln \Big(\dfrac {\sqrt a+\sqrt b}{2} \Big) < \dfrac {b\ln b - a\ln a}{b-a} -1 = \dfrac 1{b-a}\int _a^b \ln x \space dx =\dfrac 4{b-a}\int _\sqrt a^ \sqrt b y \ln y \space  dy$ 
Putting $c=\sqrt a , d=\sqrt b$ , we see that we need to prove  : 
$\ln \Big(\dfrac {c+d}{2} \Big) < \dfrac 2{d^2-c^2} \int _c^d y \ln y \space  dy$ $\space$  i.e. we need to prove  
$\Big(\dfrac {c+d}{2} \Big)\ln \Big(\dfrac {c+d}{2} \Big)< \dfrac 2{d-c} \int _c^d y \ln y \space  dy$ $\space$ 
i.e. $f\Big(\dfrac {c+d}{2} \Big) < \dfrac 2{d-c} \int _c^d f(y) dy$ , where $f(y)=y \ln y$ is a convex function . Hence the required inequality is true by Jensen's inequality ( as can be found in page 12 of http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.625.7106&rep=rep1&type=pdf )
