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 . 

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. 
