# Solve the following inequality $\frac{x-a}{x^2} + x \ge 2\left(1 - \frac{a}{x}\right)$

Solve the following inequality with a real parameter a:

$$\frac{x-a}{x^2} + x \ge 2\left(1 - \frac{a}{x}\right)$$

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

$$x \ge a.$$

• 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 $x^{2}$. Mar 21, 2021 at 13:03
• @mattos Thank you for your comment. I did not see that portion. The given is was corrected. Thank you!
– PRD
Mar 21, 2021 at 13:12
• the factorization is not correct Mar 21, 2021 at 13:23
• The result means that we can factor the left-hand side as No, as $x=1$ isn't a zero of the polynomial. Mar 21, 2021 at 14:54

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 $$x \in [1 - \sqrt{-2a} \quad , \quad 0.5]$$.