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?

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 $\color{red}{(1,\frac{5}{4})}$  &$\color{red}{(\frac{5}{3},\infty)}$
