Find m such that the following inequality:

$$\left|4x-2m-\frac{1}{2}\right| > -x^2 +2x + \frac{1}{2} - m$$

is always true for $\forall x \in R$.

1st solution:

1st case

$$4x-2m-\frac{1}{2} > -x^2 + 2x +\frac{1}{2} -m$$ $$<=>x^2+2x-m-1>0$$ $$\Leftrightarrow 1^2+(m+1)< 0$$ $$\Leftrightarrow m<- 2$$

2nd case $$4x-2m-\frac{1}{2}< -(-x^2 + 2x +\frac{1}{2} -m)$$ $$\Leftrightarrow x^2-6x+3m>0$$ $$\Leftrightarrow 3^2-3m<0$$ $$\Leftrightarrow m>3$$

2nd solution:

The inequality is the same as:


Since the left-hand side is always positive, in order for the inequality to be always true, $\frac{3}{2}-m$ has to be negative, or $m > \frac{3}{2}$

The 2 solutions give different answers, so I was quite confused

But I get more confused as Wolfram Alpha gives me the solution:

$$m > \sqrt{3} - \frac{1}{4} \text{ or } m < -\sqrt{3} - \frac{1}{4} $$

There's a high chance that Wolfram Alpha's solution is correct (after testing out some $m$ value). How do I approach their solution? (Or maybe if you believe that solution is wrong, then what's the exact solution to the problem?)


Feedback on the First Solution

Your solution is somewhat correct but not complete. In fact the question asks us to find all $m$ that the given inequality holds for all $x \in \mathbb{R}$. Your decomposition into the cases is correct, but you then tried to find all $m$ that each of the cases holds for all $x \in \mathbb{R}$, which is not a necessity to find all such $m$'s because all $x \in \mathbb{R}$ need not satisfy each cases, separately; in other words, the "all $x \in \mathbb{R}$" must lie before the decomposition.

Clarifying the above explanation, consider, for example, $m=2$. According to your conclusion, $m=2$ must not be acceptable. However, $m=2$ satisfies the inequality in the following way, for example: for $x\in [0, \infty )$ the first case holds, and for $x\in (- \infty , 0)$ the second case holds. So $m=2$ satisfies the inequality for all $x \in \mathbb{R}$.

Feedback on the Second Solution

The inequality is the same as:$$(x-1)^2+|4x-2m-\frac{1}{2}| \gt \frac{3}{2}-m$$Since the left-hand side is always positive, in order for the inequality to be always true $\frac{3}{2}-m$ has to be negative.

Your argument is not correct because for some values of $m$ the right-hand side may be positive while the left-hand side may be always greater than the right-hand side for all $x \in \mathbb{R}$ (For example, consider $m=-2$).

The Correct Solution

When you face an absolute value inequality that using some absolute value properties to solve it may lead to some misleading results, it is better first to get rid of the absolute value. So, let us solve this problem in this way as follows.$$\left |4x-2m-\frac{1}{2} \right | =\begin{cases} 4x-2m-\frac{1}{2}, & \text{if } x \ge \frac{4m+1}{8} \\ -(4x-2m- \frac{1}{2}), & \text{if } x \lt \frac{4m+1}{8} \end{cases}.$$ So, the original problem is now translated into finding all $m$ for which the following inequalities hold:$$\begin{cases} 4x-2m-\frac{1}{2} \gt -x^2+2x+\frac{1}{2}-m , & \text{if } x \ge \frac{4m+1}{8} \\ -(4x-2m- \frac{1}{2}) \gt -x^2+2x+\frac{1}{2}-m, & \text{if } x \lt \frac{4m+1}{8} \end{cases}.$$Solving each inequality, we have$$\begin{cases} x^2+2x-m-1 \gt 0, & \text{if } x \ge \frac{4m+1}{8} \\ x^2-6x+3m \gt 0, & \text{if } x \lt \frac{4m+1}{8} \end{cases}.$$As we know, a quadratic $x^2+bx+c$ is positive for all $x \in \mathbb{R}-[r_1,r_2]$, where $r_1$ and $r_2$ are the roots of the quadratic. The roots of the first quadratic are$$r_1=-1-\sqrt{m+2}, \qquad r_2=-1+\sqrt{m+2}.$$Since the first inequlity concerns the $x$'s greater than $\frac{4m+1}{8}$, we need to consider the $x$'s greater than $r_2$ to satisfy the first inequality. So, all possible $m$'s for which the first inequality holds can be obtained as follows.$$\frac{4m+1}{8}= -1+\sqrt{m+2} \quad \Rightarrow \quad 16m^2+8m-47=0$$$$\Rightarrow \quad m_1=-\frac{1}{4}-\sqrt{3}, \qquad m_2=-\frac{1}{4}+\sqrt{3}$$Only $m=m_2$ is acceptable because there are $m \gt m_1$ (for example, $m=0$) not satisfying the first inequality.

A similar argument can be applied to the second inequality. The roots of the second quadratic are$$r'_1=3-\sqrt{9-3m}, \qquad r'_2=3+\sqrt{9-3m}.$$Since the second inequlity concerns the $x$'s less than $\frac{4m+1}{8}$, we need to consider the $x$'s less than $r'_1$ to satisfy the second inequality. So, all possible $m$'s for which the second inequality holds can be obtained as follows.$$\frac{4m+1}{8}=3-\sqrt{9-3m} \quad \Rightarrow \quad 16m^2+8m-47=0$$$$\Rightarrow \quad m'_1=-\frac{1}{4}-\sqrt{3}, \qquad m'_2=-\frac{1}{4}+\sqrt{3}$$Only $m=m'_1$ is acceptable because there are $m \lt m'_2$ (for example, $m=0$) not satisfying the second inequality.

Thus, the following are the possible values for $m$ satisfying the original inequality:$$m \in (-\infty , -\frac{1}{4}-\sqrt{3} ) \cup (-\frac{1}{4}+\sqrt{3} , \infty ).$$


Your two solutions are mathematically correct, but you when you reach $m<-2$, you have calculated for which values of $m$, we have $m+2>0$ and you have not shown that $\forall m>-2, (x+1)^2>m+2, \forall x \in R$. This explain the error in the first method. For the second method, you have shown that the inequality: $$(x-1)^2+|4x-2m-\frac{1}{2}|>\frac{3}{2}-m$$ have solutions when $m<\frac{3}{2}$, but this not explain for which $m$, we have: $$\left|4x-2m-\frac{1}{2}\right| > -x^2 +2x + \frac{1}{2} - m$$ In general, when you have to solve this type of inequality, you have to solve the system: $$\left\{\begin{matrix} 4x-2m-\frac{1}{2}\geq0 \\ -x^2 +2x + \frac{1}{2} - m\geq0 \\ 4x-2m-\frac{1}{2} > -x^2 +2x + \frac{1}{2} - m \end{matrix}\right. \vee \left\{\begin{matrix} 4x-2m-\frac{1}{2}<0 \\ -x^2 +2x + \frac{1}{2} - m<0 \\ -(4x-2m-\frac{1}{2}) > -x^2 +2x + \frac{1}{2} - m \end{matrix}\right. $$ These system are quite complicated, and I will give directly the solutions: $$\left\{\begin{matrix} x>\sqrt{m+2}-1 \\ x<-\sqrt3\sqrt{3-m}+3 \end{matrix}\right.$$ When $m<-2$ and $m>3$ the square roots are not defined, so these values of $m$ are correct.

Also, we want the two expression to be the same because we want a single solution, so: $$\sqrt{m+2}-1=-\sqrt3\sqrt{3-m}+3$$ The solutions are: $$m=\sqrt3-\frac{1}{4} \vee m=-\sqrt3-\frac{1}{4}$$

When $-\sqrt3-\frac{1}{4}<m<\sqrt3-\frac{1}{4}$ there are solutions, while when: $$m > \sqrt{3} - \frac{1}{4} \lor m < -\sqrt{3} - \frac{1}{4}$$ there aren't.

  • $\begingroup$ Actually Wolfram Alpha's solution is correct. Try to solve the inequality with $m = -2$, which doesn't satisfy the 1st solution. $\endgroup$ – The 2nd Apr 7 '20 at 8:40
  • $\begingroup$ @KiênP.S.: note that in the condition $m > \sqrt{3} - \frac{1}{4} \lor m < -\sqrt{3} - \frac{1}{4}$ there is also the condition $m=-2$, so my results are correct. $\endgroup$ – Matteo Apr 7 '20 at 9:24
  • $\begingroup$ but your answer state that there aren't any solution when $m > \sqrt{3} - \frac{1}{4} \lor m < -\sqrt{3} - \frac{1}{4}$ $\endgroup$ – The 2nd Apr 7 '20 at 10:05

Your mistake in the first solution: at some stage you drop the variable $x$ by using $\forall x\in\mathbb R$. But as you are doing this case analysis, the $\forall$ no longer holds.

Your mistake in the second solution: from $a>0, a>b$, you conclude $0>b$, which is not tight. (Take $a=2,b=1$.)


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