Let $$f(z)=(1+z)^{3/4}-\left(\frac{3}{8}+\frac{\sqrt{3}}{4}\right)^{1/4}-\frac{\left(3 z+\sqrt{6} \sqrt{-1+z^2}\right)^{3/4}}{\left(2 \left(2+\sqrt{3}\right)\right)^{3/4}}.$$ Is there a simple proof that this function is positive for $z\ge 1$?



An inequality for a real function
Make the substitution $u=(1+z)^{3/4}$, so that $u\ge2^{3/4}$, $z=u^{4/3}-1$, and the inequality in question becomes
$$F(u):=f(u^{4/3}-1)>0. \tag{1}\label{1}$$


For $z=u^{4/3}-1>1$
$$F''(u)=\frac{5 u^2 \sqrt{z-1}+\sqrt{6} \left(2 z^2+3 z-1\right)}{\left(2
 \left(2+\sqrt{3}\right)\right)^{3/4} (z-1)^{3/2} (z+1) \left(\sqrt{6} \sqrt{z^2-1}+3
 z\right)^{5/4}}, $$
which is manifestly $>0$. So, $F(u)$ is convex in $u\ge2^{3/4}$. Also, for $u\_\*:=\frac{212}{100}$ we have $F'(u\_\*)=0.00227\ldots>0$. So,
$$\min\_{u\ge2^{3/4}}F(u)
\ge\min\_{u\ge2^{3/4}}[F(u\_\*)+F'(u\_\*)(u-u\_\*)] \\ 
=F(u\_\*)+F'(u\_\*)(2^{3/4}-u\_\*)=0.0579\ldots>0.$$
Thus, \eqref{1} is proved. $\quad\Box$


