Solve the following inequality with a real parameter a:
$\frac{x-a}{x^2} + x \ge 2\left(1 - \frac{a}{x}\right)$
I am not sure if my answer is correct. Please feel free to share your thoughts.
First, we need to multiply both sides of the given inequality by the LCD, which is $x^2$. Since $x^2$ is always positive for all real numbers $x$, we don't have to change the direction of the inequality.
$x^2 \left[\frac{x-a}{x^2} + x \right] \ge 2x^2 \left(1 - \frac{a}{x}\right)$
$x-a+x^3 \ge 2x^2 - 2ax$
Isolate all terms on the left-hand side of the inequality.
$x^3-2x^2+(2a+1)x - a \ge 0$
Factor the left-hand side of the inequality. Since $a$ is a constant, it will always have a factor of $1$. Let us try to determine if $x-1$ is a factor using synthetic division.
$\begin{array}{cccccc} 1| & 1 & -2 & 2a+1 & -a \\ & & 1 & -1 & 2a \\ \hline & 1 & -1 & 2a & a \end{array}$
The result means that we can factor the left-hand side as
$(x-1) \left[x^2-(a+1)x+a\right] \ge 0.$
We can further factor the quadratic term on the left-hand side.
Note that we can express the inequality as
$(x-1) \left[x^2+(-a-1)x+a\right] \ge 0.$
The middle term is equal to the sum of $-a$ and $-1$, and the last term can be expressed as $(-a) * (-1)$. Hence, we can further express this as
$(x-1)(x-1)(x-a) \ge 0$
$(x-1)^2(x-a) \ge 0.$
Notice that for all real values of $x$, $(x-1)^2$ will always be positive. So, to make the left-hand side always greater than or equal to 0, the factor $x - a$ should be greater than or equal to $0$. That is $(x-a) \ge 0.$
Therefore, we can now say that the solution to the inequality must be
$x \ge a.$