略

$f^{\prime}(x)=\dfrac{a}{x}+\dfrac{1}{2\sqrt{1+x}}$，当 $a=-\dfrac{3}{4}$ 时，$f^{\prime}(x)=-\dfrac{3}{4x}+\dfrac{1}{2\sqrt{1+x}}$，令 $x=\tan^2{t},t\in\left(0,\dfrac{\pi}{2}\right)$，
则 $s=-\dfrac{3}{4x}+\dfrac{1}{2\sqrt{1+x}}=-\dfrac{\cos{t}(2\cos{t}-1)(\cos{t}+2)}{4\sin^2t}$，
容易得到当 $\cos{t}\in(0,\dfrac{1}{2}),此时x=\tan^2t\in(3,+\infty)$，$s>0$，综合可知 $f(x)$ 递增区间为 $(3,+\infty)$，递增区间为 $(0,3)$．

略

令 $b(x)=a\ln{x}+\sqrt{1+x}-\dfrac{\sqrt{x}}{2a}$，命题成立的必要条件为 $b(1)\le0$，解得 $a\in\left(0,\dfrac{\sqrt2}{4}\right]$．
于是命题转化为 $g(a)=2a^2\ln{x}+2a\sqrt{1+x}-\sqrt{x}\le0$ 在 $x\in\left[\dfrac{1}{{\rm{e}}^2},+\infty\right)$ 恒成立，在 $a\in\left(0,\dfrac{\sqrt2}{4}\right]$ 情况下求 $a$ 的范围．接下来进入二次函数讨论．
1°当 $x=1$，即 $2\ln{x}=0$，命题已经验证，$a\in\left(0,\dfrac{\sqrt2}{4}\right]$．
2°当 $0<x<1$，即 $2\ln{x}<0$，
考虑 对称轴 $l(x)=-\dfrac{\sqrt{1+x}}{2\ln{x}}$，
以及判别式 $\Delta(x)=4(1+x+2\sqrt{x}\ln{x})=4\sqrt{x}\left(\dfrac{1}{\sqrt{x}}+\sqrt{x}+4\ln{\sqrt x}\right)$，
2．1°考虑函数 $h(x)=\dfrac{1}{x}+x+4\ln{x}$，
$h^{\prime}(x)=-\dfrac{1}{x^2}+1+\dfrac{4}{x}=\dfrac{x^2+4x-1}{x^2}>0(x>\dfrac{1}{e})$，
$h(\dfrac{1}{e})<0,h(\dfrac{1}{2})=2.5-4\ln2<0$，
即当 $x\in\left[\dfrac{1}{e^2},\dfrac{1}{4}\right]$，$\Delta(x)<0$，符合题意．
2．2°而当 $1>x>\dfrac{1}{4}$ 时，由 $l(x)=\sqrt{1+x}\left(-\dfrac{1}{2\ln x}\right)$ 两个恒正单调增函数相乘，是个递增函数．
$l(x)>l(\dfrac{1}{4})=\dfrac{\sqrt{5}}{8\ln2}>\dfrac{\sqrt{2}}{4}$，当 $a\in\left(0,\dfrac{\sqrt2}{4}\right]$，$g(a)\leqslant g(\dfrac{\sqrt{2}}{4})$
令 $w(x)=g(\dfrac{\sqrt{2}}{4})=\dfrac{1}{4}\ln x+\dfrac{\sqrt{2}}{2}\sqrt{1+x}-\sqrt{x}$
$w^{\prime}(x)=\dfrac{1}{4x}+\dfrac{\sqrt{2}}{4\sqrt{1+x}}-\dfrac{1}{2\sqrt{x}}$，
三角换元令 $x=\tan^2t$，其中 $t$ 是锐角，在本次讨论中，$\dfrac{1}{4}<\tan t<1$．
上式变形为 $\dfrac{\cos{t}(\cos{t}+\sqrt{2}\sin^2t-2\sin{t})}{4\sin^2t}$，
令 $v(t)=\cos{t}+\sqrt{2}\sin^2t-2\sin{t}$，$v^{\prime}(t)=-\sin{t}+2\sqrt{2}\sin{t}\cos{t}-2\cos{t}<-\sin^2{t}+2\sqrt{2}\sin{t}\cos{t}-2\cos^2{t}\le0$，故 $v(t)$ 单调递减，又 $v(\dfrac{\pi}{4})=0$，因此 $\dfrac{1}{4}<\tan t<1$ 时，$w^{\prime}(x)>0$，$w(x)$ 单调递增，$w(x)<w(1)=0$，$a$ 的范围依然是 $a\in\left(0,\dfrac{\sqrt2}{4}\right]$
3°当 $x>1$ 时，$l(x)<0$，$g(0)=-sqrt{x}<0$．$g(\dfrac{\sqrt{2}}{4})$ 依然讨论 $w(x)$ 此时范围为 $\tan{t}>1$，$w^{\prime}(x)<0$，结合 $w(1)=0$．
综上 $a\in\left(0,\dfrac{\sqrt2}{4}\right]$．