$\cos{y}=x+\frac{1}{x}$, possible for any values of $y$? If for real values of x, $\cos{y}=x+\frac{1}{x}$,
Then how do we prove that no value of $y$ is possible ?
 A: The value of $\cos(y)$ is between $-1$ and $1$.  Therefore, for this equality to hold, $x+\frac{1}{x}$ must be between $-1$ and $1$.


*

*If $x>1$, then $x+\frac{1}{x}>x>1$.

*If $x=1$, then $x+\frac{1}{x}=2>1$.

*If $0<x<1$, then $\frac{1}{x}>1$, so $x+\frac{1}{x}>\frac{1}{x}>1$.

*If $x=0$, then $x+\frac{1}{x}$ is not defined.

*If $-1<x<0$, then $\frac{1}{x}<-1$ so $x+\frac{1}{x}<\frac{1}{x}<-1$.

*If $x=-1$, then $x+\frac{1}{x}=-2<-1$.

*If $x<-1$, then $x+\frac{1}{x}<x<-1$.
Since $x+\frac{1}{x}$ is either greater than $1$ or less than $-1$, the equality can never hold (at least in the reals).
Alternately, one could use calculus and compute minima and limits as $x\rightarrow\pm\infty$ and $x\rightarrow 0^{\pm}$.  However, since the necessary inequality can be proved with only elementary facts, I chose this method.
A: HINT consider the function $f(x) = (x + \frac{1}{x})^2$. What is lowest value of this function? What does this say about the smallest value of $|x + \frac{1}{x}|$? What then would this imply about the possibility of solving for $\cos y = x + \frac{1}{x}$?
A: Arithmetic mean is always greater than geometric mean if both number are positive or both are negative. First consider x as positive . 
Take the numbers x and 1/x their GM is 1 we get inequality as sum of x and its reciprocal as greater than or equal to 2.
If x is negative we would get it less than minus two similarly and in both cases it is beyond range of cosy which is between 1 and -1. Hence y cannot be real. Hope this helps. 
Actually the inequality is only for non negative numbers but for x as negative take y = -x and then use inequality and put back x you will get the result see this
https://en.m.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means
A: HINT:
We have  $$x^2-(\cos y)x+1=0$$
The discriminant $=\cos^2y-4=1-\sin^2y-4<0$ for all real $y$
A: For $x \ne 0$ we have:
$|x+1/x| \ge 2 \iff x^2+2+1/x^2 \ge 4\iff x^2-2+1/x^2 \ge 0 \iff (x-1/x)^2 \ge 0$.
A: For any non zero real $x$ we have $|x+1/x|=|x|+|1/x|$ (because both terms have same sign). If $|x| \ge 1$ then it follows that $|x+1/x|>1$. As, for any real $y$, $|\cos(y)| \le 1$, the equality is impossible. Similarly, if $|x|<1$, $x \neq 0$, then $|1/x|>1$ and we have the same conclusion.
