The given inequality
$$g(u)=6 + 5 \sin u + \sin(2 u)- \cos u - \cos(2 u) \ge 0$$
for $u\in\left[-\frac \pi 2, \frac \pi 2\right]$, comes out from an answer given to this other recent question.
The solution given there makes use of elementary analysis tools, mainly derivatives, and some numerical check.
Can we find an alternative and more effective or elegant proof?
We have \begin{align*} &6 + 5 \sin u + \sin(2 u)- \cos u - \cos(2 u)\\ \ge{}& 6 + 5 \sin u + (-1)- 1 - \cos(2 u)\\ ={}& 3 + 5\sin u + 2\sin^2 u\\ ={}& (2\sin u + 3)(\sin u + 1)\\ \ge{}& 0 \end{align*} where we use $\cos (2u) = 1 - 2\sin^2 u$.
Express the trigonometric functions in $t=\tan(\frac u2)$ and eliminate the common denominator $(1+t^2)^2.$ The resulting numerator is the 4-th degree polynomial $$P(t)=7t^4+6t^3+20t^2+14t+4$$ This can be proven by elementary means to be always positive, for example by noting that it is equal to $$6t^2(t^2+t+1)+t^4+(\frac 7 2t+2)^2+\frac7 4t^2$$
