Problem related to Inequalities. First I would give the background of this question. The question was to prove that if $x$ is real, $$ \sin\theta =x+\frac{1}{x} $$ is impossible. I thought that we can prove this impossible, if we can prove $ x + \frac {1}{x} $ as either less than -1 or greater than 1 ( outside the range of $\sin\theta$ for any $\theta$ ).
I tried to prove two inequalities for this, which were - $$ x+\frac{1}{x}>1$$ if $x>0$ and $x$ is real.
And $$ x+\frac{1}{x}<1$$ if $x<0$ and $x$ is real.
Solving first inequality gave $$ x^2-x+1>0 $$ and then subtracting x from both the sides of inequality gave $$ ( x-1)^2>- x $$ But the square at R.H.S is grater than 0 for any real x ( even if they are negative). 
So I have proved by this that $ x+\frac{1}{x}>1$ for even negative real $x$. 
But if I take any negative numeric value, then $ x+\frac{1}{x}$ is negative .
So my question is why I got this result true for negative real $x$ also even though it is not true ?
Please help and thanks.
 A: You did not take into account that multiplying an inequality by
a negative number reverses the inequality sign. So
$$
 x+\frac{1}{x}>1 \Longleftrightarrow x^2 - x + 1 > 0
$$
holds only if $x > 0$, and for negative $x$ you would have
$$
 x+\frac{1}{x}>1 \Longleftrightarrow x^2 - x + 1 < 0 \, .
$$
A: If $x<0,$ let $x=-y$
$$x+\dfrac1x=-\left(y+\dfrac1y\right)$$
As $y>0,\dfrac{y+\dfrac1y}2\ge\sqrt{y\cdot\dfrac1y}$
Alternatively, $$x^2-x\sin\theta+1=0$$
As $x$ is real, the discriminant must be $\ge0$
A: The function $f(x)=x+\frac{1}{x}$ has two critical points: indeed
$$f'(x) = 1-\frac{1}{x^2} =0 \iff |x|=1$$
So that the critical points are $y=+1$ and $z=-1$. 
Moreover, by looking at the sign of $f'$, note that $f$ is increasing on $(-\infty,-1)$ and $(1,\infty)$ and $f$ is decreasing on $(-1,0)$ and $(0,1)$ implying that $y=1$ is a local minimum and $z=-1$ is a local maximum. 
As $f(1)=2$ and $f(-1)=-2$ it follows that $f(x)\geq 2$ for $x>0$ and $f(x)\leq -2$ for $x<0$, that is $|f(x)|\geq 2$ for every $x$ and so $f(x)=\sin(\theta)$ never has a solution.
