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

$x \ge a.$

  • 1
    $\begingroup$ As far as I can tell, you should have quartic terms $x^{4}$ in the expression; it seems you never multiplied the RHS of the inequality through by $x^{2}$. $\endgroup$ Mar 21, 2021 at 13:03
  • $\begingroup$ @mattos Thank you for your comment. I did not see that portion. The given is was corrected. Thank you! $\endgroup$
    – PRD
    Mar 21, 2021 at 13:12
  • 2
    $\begingroup$ the factorization is not correct $\endgroup$
    – Lozenges
    Mar 21, 2021 at 13:23
  • $\begingroup$ The result means that we can factor the left-hand side as No, as $x=1$ isn't a zero of the polynomial. $\endgroup$ Mar 21, 2021 at 14:54

1 Answer 1


Your argument is fine up to the point $$ f(x) = x^3-2x^2+(2a+1)x - a \ge 0 $$ Then, your roots are wrong. So one needs a curve analysis. We have $$ f´(x) = 3x^2-4x+(2a+1) $$ so extreme points are $x = \frac13(2 \pm \sqrt{1 - 6 a})$. So for $a > \frac16$, there are no extremal points at all, which means $f(x)$ is rising everywhere. Since it is ugly to determine the (sole) root of $f(x)$ in this case, we can make a nice estimate. Notice $$ f(x) = x(x-1)^2 +a(2x -1) $$ which lets us give the bound that $f(x)$ is $\ge 0$ at least for $a \ge 0$ and $x\ge 0.5$. In fact, for large $a$, $x\ge 0.5$ becomes asymptotically exact.

This bound $x\ge 0.5$ is not that bad also for smaller positive $a$. The true values are that for $a = \frac16$ we have $x \ge 0.161$, and for $a = 0$ we obviously have $x \ge 0$.

Considering negative $a$, the behavior of $f(x)$ becomes more beasty. Taking the above extremal values gives us that the value of $f(x)$ at the minimum is $\frac{a}{3} + \sqrt{1-6a} \frac{12 a -2}{27}+ \frac{2}{27} \le 0$ and at the maximum, $\frac{a}{3} - \sqrt{1-6a} \frac{12 a -2}{27}+ \frac{2}{27} \ge 0$. Clearly, our third order polynomial will therefore have exactly one interval where $f(x) \ge 0$ about the maximum $\frac13(2 - \sqrt{1 - 6 a})$, and then $f(x)$ will stay positive again for $x > x_0$, where that $x_0$ can be estimated for negative $a$: $$ f(x) = x(x-1)^2 +a(2x -1) \ge x(x-1)^2 +a(2x) = x [(x-1)^2 +2a] $$ Hence $f(x) \ge 0$ at least for $x \ge x_0 = 1 + \sqrt{-2a}$ which indeed is a rather precise approximation. The same bound on $f(x)$ can be used to say, for $a < -\frac18$, that $f(x) \ge 0$ at least in the interval $x \in [1 - \sqrt{-2a} \quad , \quad 0.5]$. That holds true, as it is known that $f(x) \ge 0 $ on both ends of that interval, and since we know from the above discussion that there is only one continuous interval with positive $f(x)$ about the maximum. Again, this is a rather precise approximation for more negative $a$.

Let's summarize these findings: We have, with rather precise approximation, that $f(x) \ge 0$ at least for:
$a \ge 0$ and $x\ge 0.5$;
$a \le 0$ and $x \ge 1 + \sqrt{-2a}$ ;
$a \le -\frac18$ and $x \in [1 - \sqrt{-2a} \quad , \quad 0.5]$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .