Difficult inequality with three real variables For any real $e, t, \sigma$ such that
\begin{aligned}
\label{s}
     0&<e<1\,,\\
     0&<t<\pi\,,\qquad\qquad\qquad(1)\\
-\pi/2&\leqslant\sigma\leqslant\pi/2
\end{aligned}
the inequality
\begin{aligned}
(1-e^2\cos^2\sigma)\eta&+(1+e\cos(t-\sigma))(e\eta\cos\sigma\cos t
                                                         +e\sin t)>\\
                       &>(1+e\cos(t-\sigma))\sqrt{1-(e\eta\cos\sigma\sin t-
                                                    e\cos t)^2}\,,\qquad(2)
\end{aligned}
where $\eta=\sqrt{1-e^2}$, holds.
I have "proved" (2) numerically by three-dimensional brute-force method over the set (1). I used a three-dimensional grid with a high partition density in my C++ program. So I am sure that (2) holds on (1). Moreover, I proved (2) analitically for special case $\sigma=-\pi/2$.
In order to prove (2) analitically, I suppose that one have to do as follows. Denote for shortness
\begin{aligned}
\tau&=1-e^2\cos^2\sigma\,,\\
\gamma&=1+e\cos(t-\sigma)\,,\\
\delta_1&=\eta\cos\sigma\cos t+\sin t\,,\\
\delta_2&=\eta\cos\sigma\sin t-\cos t\,.
\end{aligned}
It is easy to prove that the expression under the root in (2) is positive on (1). Further, suppose that the left-hand side of (2) is also positive (at least nonnegative) on (1). Then (2) holds if and only if the square of the left-hand side of (2) is greater than the square of its right-hand side. After squaring both sides of (2) and noticing that
$$
\delta^2_1+\delta^2_2=1+\eta^2\cos^2\sigma
$$
one obtains after some computations the following inequality
\begin{equation}
\frac{\tau\eta}{\gamma^2}+\frac{2e\delta_1}{\gamma}>\eta\qquad\qquad(3)
\end{equation}
that has to be proved. So if the left-hand side of (2)
\begin{equation}
\tau\eta+e\gamma\delta_1\geqslant0\,\qquad\qquad\qquad(4)
\end{equation}
on (1), it only remains to establish the validity of (3) on (1).
In my opinion the task (1), (3), (4) is simpler than the original task (1), (2), but I am stuck at this stage. Maybe there are some other ways to deal with (1), (2), for example without the squaring (2)? Any ideas?

Continuity of the functions
\begin{align}
f(e, t, \sigma) &\stackrel{\mathrm{def}}{=} \mathrm{LHS}(2) =
                                            \tau\eta+e\gamma\delta_1,\\
g(e, t, \sigma) &\stackrel{\mathrm{def}}{=} \mathrm{RHS}(2) =
                                            \gamma\sqrt{1-e^2\delta_2^2},\\
h(e, t, \sigma) &\stackrel{\mathrm{def}}{=} f(e, t, \sigma)-g(e, t, \sigma)
\end{align}
on the set (1) completes the proof. We want to show that $h(e, t, \sigma)>0$ on (1).
Suppose that there is a point $(e_0, t_0, \sigma_0)\in(1)$ such that
$h(e_0, t_0, \sigma_0)=0$. Then
$$
  f(e_0, t_0, \sigma_0)=g(e_0, t_0, \sigma_0)\Rightarrow
  \bigl[f(e_0, t_0, \sigma_0)\bigr]^2=\bigl[g(e_0, t_0, \sigma_0)\bigr]^2.
$$
Contradiction, because as we already know,
$$
\bigl[f(e, t, \sigma)\bigr]^2>\bigl[g(e, t, \sigma)\bigr]^2
$$
everywhere in (1). By continuity of $h(e, t, \sigma)$ it means that
$h(e, t, \sigma)$ is either positive or negative on (1). Since we saw that
there are subsets of (1) where $h$ is positive, we conclude that
$h(e, t, \sigma)>0$ everywhere in (1).
