Prove that if $x \in \mathbb R$ and $x>2$ then $y+ \frac{1}{y} = x$ will have real solution. 
Prove that for real number $x$, if $x > 2$ then there is real number $y$ such that $y + \frac{1}{y} = x$

My attempt: 
Rewriting equation, we have:
$$\tag1y + \frac{1}{y} = x$$
$$\tag2 \frac{y^2+ 1}{y} = x$$
$$\tag3 y^2+ 1 = yx$$
$$\tag4 y^2 - yx + 1 = 0$$
We know that for quadratic equations of the format $aq^2 + bq + c$ ($a,b,c$ are constants and q is variable), if discriminant is positive, then equation will have real solutions.
Let discriminant be denoted by $D$, then
$D = \sqrt{b^2- 4ac}$
Substituting our values into the equation, we have:
$D = \sqrt{x^2 - 4}$
It can be seen that if $x > 2$ then $D$ is positive, hence $y + \frac{1}{y} = x$ will have real solution.
Is it correct?
 A: This is essentially correct. One small thing which is missing: you should say that the equation $y^2-xy+1$ has non-zero real solutions, since the step where you multiplied both sides by $y$ could in principle have added an extra "solution" $y=0$ which was not a solution of the original equation. But that has not happened in this case since clearly $y=0$ is not a solution of your quadratic.
A: Let $y >0$; $x>2$;
AM-GM:
$y+1/y \ge 2\sqrt{y\cdot 1/y}=2$;
Equality for $y=1/y$, i.e $y=1$;
Consider $f(y):=y+1/y$, $y \in [1,\infty)=:D$.
$f([1,\infty))= [2,\infty)$.
$f$ is continuos in its domain $D$, 
MVT:
For given $x \in (2, \infty)$ there is a $y_0$ with
$f(y_0)=x$; 
Hence a real solution.
Note: $f(1/y_0)=x$, is also a solution.
A: Multiplying both sides by $y$ and putting the result in standard form, we have
$$y^{2} - xy + 1 = 0.$$
Hence, the discriminate of this quadratic equation is $x^{2} - 4$; and so, if $x > 2$, we certainly get two distinct real solutions. But also, if $x < -2$ we may draw the same conclusion as well. Finally, if $x = \pm 2$, then we get exactly one real solution.
