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?
 A: I think you chose a right way. I followed it and proved inequality (3). I failed to prove inequality (4), but we also need it to be true because the right hand side of inequality (2) is positive. 
We can prove (3) as follows. Since  $\gamma>0$, we have to prove 
$$\tau\eta>\gamma(\eta\gamma-2e\delta_1)$$ 
$$(1-e^2\cos^2\sigma)\eta>(1+e\cos t\cos\sigma+e\sin t\sin\sigma)(\eta(1+e\cos t\cos\sigma+e\sin t\sin\sigma)-2e(\eta\cos\sigma\cos t+\sin t))$$
After routine transformations we obtain 
$$0>\eta e\sin^2t+2\eta\sin t\sin\sigma-2\sin t-2e\sin t\cos t\cos\sigma-2e\sin^2 t\sin\sigma$$
Since $\sin t>0$, this is equivalent to 
$$0>\eta e\sin t+2\eta\sin\sigma-2-2e\cos t\cos\sigma-2e\sin t\sin\sigma$$
We show that 
$$2>\eta e\sin t+2\eta\sin\sigma-2e\cos t\cos\sigma-2e\sin t\sin\sigma.$$
By Cauchy-Schwarz inequality, $$(\eta-e\sin t)\sin\sigma-e\cos t\cos\sigma\le\sqrt{(\eta-e\sin t)^2+(e\cos t)^2}=\sqrt{1-2\eta e\sin t}.$$
Put $x=\eta e\sin t$. Then $0<x\le \frac{\eta^2+e^2}2=\frac 12$ and we have to check that
$$2>x+2\sqrt{1-2x}$$
$$(2-x)^2>4(1-2x)$$
$$x+4>0.$$
A: Many thanks for your answer, Alex. I suppose that (4) is harder than (3) to be dealt with and therefore I hope that the most difficult part of the task thanks to your efforts is overcome. As for the rest task (1, 4) here is my simple thoughts about it.
First of all, if $0<t\leqslant\pi/2$, then $\cos t\geqslant0$ and (4) obviously holds on (1). Thus putting $\varphi=t-\pi/2$ the inequality (4) takes the form
$$
\tau\eta+(1+e\sin(\sigma-\varphi))(e\cos\varphi-e\eta\cos\sigma\sin\varphi)>0.\qquad (5)
$$
We have to verify that (5) is valid on the set
\begin{align}
0 &< e < 1,\\
0 &< \varphi < \pi/2,\qquad\qquad\qquad(6)\\
-\pi/2 &\leqslant\sigma\leqslant\pi/2.
\end{align}
Further, we can prove (5, 6) for every $0<e<1/2$ as follows. We obviously have
\begin{align}
(1+e\sin(\sigma-\varphi))\leqslant 1+e,\\
(e\cos\varphi-e\eta\cos\sigma\sin\varphi)\geqslant-e\eta.
\end{align}
Thus
$$
(1+e\sin(\sigma-\varphi))(e\cos\varphi-e\eta\cos\sigma\sin\varphi)\geqslant-e\eta(1+e)
$$
This is rough (not exact) low bound for $e\gamma\delta_1$ in (4). So
\begin{align}
\tau\eta-e\eta(1+e) &> 0\\
               \tau &> e(1+e)\\
                    1 &> e+e^2(1+cos^2\sigma)\\
                    1 &> e+2e^2\\
             2e^2+e-1 &< 0.
\end{align}
The most frustrating thing here is that the quadratic polynomial $2e^2+e-1$ is negative only for $0\leqslant e<1/2$.
So the task (1, 2) can be reduced to the proving the validity of (5) on the set
\begin{align}
1/2 &\leqslant e < 1,\\
0 &< \varphi < \pi/2,\qquad\qquad\qquad(7)\\
-\pi/2 &\leqslant\sigma\leqslant\pi/2.
\end{align}
I think that one has to introduce a new variable $s=1-e$ or somthing like that. Ironically, from the geometrical content of this problem it clearly follows that
the more e, the bigger difference between the right and the left hand side of (2)
should be. That difference must decrease when decrease $\sin t$. And when $\sin t=0$
the left hand side of (2) is equal to the its right hand-side.
I thought about more poweful tools, like expanding these quantities in Maclaurin series in e, or trigonometric series, and investigating their coefficients. Maybe one ought to try use complex representation on trigonometric expressions.
A: 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).
