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?

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 $x \in (1, \frac54) \cup (\frac53, \infty)$. 
