How to prove that the following expression is positive for $u\ge 0$ and $q\in(0,\pi/3)$. $$\frac{2 \sqrt{2} ((2+u) (1+\cos(q)))^{3/4}}{3^{3/4}}-\frac{2^{3/4} (1+\sec(q))}{\sqrt{3} \sec(q)^{3/4}}-\left(1+u+\sqrt{-1+(1+u)^2} \sqrt{1-\frac{4 \sin(q)^2}{3}}\right)^{3/4}$$



An inequality involving 2 variables
Letting $c:=\cos q$, we have $c\in[1/2,1]$. Writing $\sec q=1/c$ and $\sin^2q=1-c^2$, and multiplying the big expression by $3^{3/4}c^{1/4}$, we see that the inequality in question can be rewritten as
\begin{equation\*}
 f(c,u)\overset{\text{(?)}}>0, \tag{10}\label{10}
\end{equation\*}
where $(c,u)\in[1/2,1]\times[0,\infty)$,
\begin{equation\*}
 f(c,u):=\big(1-(r\_1(c,u)+r\_2(c,u))\big)\,2\sqrt2\, (c (1 + c)^3 (2 + u)^3)^{1/4}, 
\end{equation\*}
\begin{equation\*}
 r\_1(c,u):=\frac{\sqrt[4]{3}}{2^{3/4}}\, \sqrt[4]{\frac{c+1}{c (u+2)^3}},\quad 
 r\_2(c,u):=\frac1{2 \sqrt{2}}\,
 \left(\frac{\sqrt{3} \sqrt{\left(4 c^2-1\right) u (u+2)}+3 u+3}{c u+2
 c+u+2}\right)^{3/4}. 
\end{equation\*}


So, it is enough to show that
\begin{equation\*}
 r\_1(c,u)+r\_2(c,u)\overset{\text{(?)}}<1. \tag{20}\label{20}
\end{equation\*}


Note that $r\_1(c,u)$ is decreasing in $c\in[1/2,1]$ and in $[0,\infty)$. So,
\begin{equation\*}
 r\_1(c,u)\le r\_1(1/2,73/10)<1-\frac{3^{3/4}}{2 \sqrt{2}}\quad
 \text{for }(c,u)\in[1/2,1]\times[73/10,\infty). 
\end{equation\*}
Next, for all $(c,u)\in[1/2,1]\times[0,\infty)$
\begin{equation\*}
\begin{aligned}
 &\big((2\sqrt2\,r\_2(c,u))^{4/3}-3\big) \\ 
 &\times (c+1) (u+2) \left(\sqrt{3} \sqrt{\left(4 c^2-1\right) u (u+2)}+3 c u+6 c+3\right) \\ 
 &=\left(3-3 c^2\right) u^2+6 \left(2 c^2+3 c+1\right) u+9 (2 c+1)^2,
\end{aligned}
\end{equation\*}
which is manifestly $>0$, whence $r\_2(c,u)<\dfrac{3^{3/4}}{2 \sqrt{2}}$.
So, we have \eqref{20} and hence \eqref{10} for $(c,u)\in[1/2,1]\times[73/10,\infty)$.


It remains to prove \eqref{10} for $(c,u)\in[1/2,1]\times[0,73/10]$.


To do this, note that
\begin{equation}
 f(c,u)=f\_1(c,u)-f\_2(c,u), 
\end{equation}
where
\begin{equation}
 f\_1(c,u):=2 \sqrt{2} \sqrt[4]{c (c+1)^3 (u+2)^3}-2^{3/4} \sqrt[4]{3} (c+1), 
\end{equation}
\begin{equation}
 f\_2(c,u):=\sqrt[4]{c \left(\sqrt{3} \sqrt{\left(4 c^2-1\right) u (u+2)}+3 u+3\right)^3}. 
\end{equation}
Clearly, $f\_2(c,u)$ is increasing in $c\in[1/2,1]$ and in $[0,\infty)$.


Also clearly, $f\_1(c,u)$ is increasing in $u\in[0,\infty)$.
Also,
\begin{equation}
 g\_1(c,u):=\partial\_c f\_1(c,u)=\frac1{\sqrt{2}}
 \Big(\frac{(4 c+1) (u+2)^{3/4}}{\sqrt[4]{c^3 (c+1)}}-2 \sqrt[4]{6}\Big)
 \ge g\_1(c,0)
\end{equation}
and
\begin{equation}
 \partial\_c g\_1(c,0)=-\frac{3}{2\ 2^{3/4} c^{7/4} (c+1)^{5/4}}<0. 
\end{equation}
So, $g\_1(c,u)\ge g\_1(c,0)\ge g\_1(1,0)=2.78\ldots>0$. So, $f\_1(c,u)$ is increasing in $c\in[1/2,1]$.
So, $f\_1(c,u)$ is increasing in $c\in[1/2,1]$ and in $[0,\infty)$.


So, "partitioning" the rectangle $[0,73/10]$ into $20\times200=4000$ congruent rectangles, with vertices $(c\_i,u\_j):=(\frac{1}{2}+\frac{1}{2}\frac{i}{20},\frac{73}{10}\frac{j}{200})$,
for all $(c,u)\in[1/2,1]\times[0,73/10]$ we get
\begin{equation}
\begin{aligned}
f(c,u) 
\ge\min\big\{f\_1 & \big(c\_{i-1},u\_{j-1}\big)
 -f\_2\big(c\_i,u\_j\big) \colon \\ 
 &i=1,\dots,20,\,j=1,\dots,200\big\}=0.0247\ldots>0. 
\end{aligned}
\end{equation}
This completes the proof of \eqref{10}. $\quad\Box$


