If $b>a>0$ , how to prove that $\left(\frac {\sqrt a +\sqrt b}{2}\right )^2 < \frac 1e \left(\frac {b^b} {a^a}\right)^{\frac 1{b-a}}$?

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.

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 )