Solve the Inequality $x+\frac{x}{\sqrt{x^2-1}} \gt \frac{35}{12}$ Solve the Inequality $$x+\frac{x}{\sqrt{x^2-1}} \gt \frac{35}{12}$$
First of all the Domain of LHS is $(-\infty \:\: -1) \cup (1  \:\: \infty)$
So i assumed $x=\sec y$ since Range of $\sec y$ is $(-\infty \:\: -1) \cup (1  \:\: \infty)$
So
$$\sec y+ |\csc y| \gt \frac{35}{12}$$
Any help here to proceed?
 A: HINT:
Clearly, we need $x>0$  so will be $\sec y,\csc y\implies0< y<\dfrac\pi2$
Now $\sec y+\csc y>\dfrac{35}{12}$
$\iff\left(\dfrac{35}{12}\right)^2<\sec^2y+\csc^2y+2\sec y\csc y=\sec^2y\csc^2y+2\sec y\csc y$  as $\sec^2y\csc^2y=\sec^2y+\csc^2y$
Set $\sec y\csc y=u$ to find $$u^2+2u>\left(\dfrac{35}{12}\right)^2\iff(u+1)^2>\left(\dfrac{37}{12}\right)^2$$
As $u>0,$  $$u+1>\dfrac{37}{12}\iff\dfrac{25}{12}<u=\dfrac2{\sin2 y}\iff\dfrac{24}{25}>\sin2y=\dfrac{2\tan y}{1+\tan^2y}$$
Can you find the range of $\tan y?$
A: The inequality is:$$\frac{x}{\sqrt{x^2-1}} > \frac{35}{12} - x$$
For $x > 1$, LHS is positive and so is RHS unless $x \geqslant \frac{35}{12}$.  But if $x \geqslant \frac{35}{12}$ the inequality is trivially true. So we need to consider $x \in (1, \frac{35}{12})$, where we may square and simplify$^\dagger$ to get $(1, \frac54) \cup (\frac53, \infty)$.
Similarly, for $x< -1$, it is easy to see the LHS is negative, while the RHS is positive, so no solutions.

P.S. $\dagger$  details of the simplification asked for below...  squaring leads to 
$$\frac{x^2}{x^2-1} > \left(\frac{35}{12} - x \right)^2 \implies (12x^2-35x-49)(4x-5)(3x-5)< 0$$
The quadratic factor has no root in $(1, \frac{35}{12})$, so we look at the sign of the polynomial in $(1, \frac54), (\frac54, \frac53), (\frac53, \frac{35}{12})$ to see the solution is $x \in (1, \frac54) \cup (\frac53, \infty)$. 
A: We need only concerntrate on the interval $(1,\infty)$ ( the expression is clearly negative in the other interval) ... now change the inequality into an equation, in order to find the points where the expression will change sign
\begin{eqnarray*}
x+\frac{x}{\sqrt{x^2-1}}=\frac{35}{12} \\
\frac{x^2}{(x^2-1)}= \left( \frac{35}{12} -x \right)^2 \\
144 x^2 =(35-12x)^2(x^2-1) \\
144x^4-840x^3+937x^2+840x-1225=0 \\
(12x^2-35x-49)(4x-5)(3x-5)=0
\end{eqnarray*}
So we need to consider the intervals $(1,\frac{5}{4}),(\frac{5}{4},\frac{5}{3}),(\frac{5}{3},\infty)$
So the intervals where the inequality is satisfied is $\color{red}{(1,\frac{5}{4})}$  &$\color{red}{(\frac{5}{3},\infty)}$
