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I came across this problem:

$$\sqrt{(x^2+1)}-2x+1>0$$

So I wanted to solve it by squaring the inequality:

$$\sqrt{(x^2+1)}>2x-1$$

If I understand correctly if I want to square the equation, I need to show that $2x-1$ is a positive number, as $\sqrt{(x^2+1)}$ is positive for all real numbers.

So I got $2x-1\geqslant 0$ , so we get: $x\geqslant\frac{1}{2}$

I used this as a perimeter for the first equation when inequalities do not change, after squaring both sides I got: $$(x^2+1)>4x^2-4x+1$$ $$0>x(3x-4)$$ Then we can see that the equation is less than zero on interval: $(0,\frac{4}{3})$

Here I used intersection with $x\geqslant\frac{1}{2}$.

So the parameter becomes(Solution 1): $[\frac{1}{2},\frac{4}{3})$

Then I created another parameter for when $x<\frac{1}{2}$

After some algebra with the same square but the change of the direction of inequality, I got the new parameter intersection (Solution 2) with $x<\frac{1}{2}$ :

$$x<0$$

By joining the solution 1 and solution 2 with union, I got parameter for $x$:

$$(-\infty,0) \cup [\frac{1}{2}, \frac{4}{3}) $$

There seems to be something that I missed, as the correct solution is: $x<\frac{4}{3}$

Where did my problem-solving go wrong?

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  • $\begingroup$ It's not clear why you have the $x<0$ statement. That's where the error is, but you did not say how you got it. $\endgroup$
    – Andrei
    Oct 18, 2019 at 21:14
  • $\begingroup$ I changed around the inequalities and used intersection with the same formula, just that I flipped the inequality sign around. That's how I got $ x<0$ $\endgroup$
    – VLC
    Oct 18, 2019 at 21:16
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    $\begingroup$ When $x\le 1/2$ the right hand side is less or equal to zero. The left hand side is greater than 1, so the inequality is always true. You don't need to do any more manipulations (which might yield wrong results) $\endgroup$
    – Andrei
    Oct 18, 2019 at 21:21
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    $\begingroup$ "I need to show that 2x−1 is a positive number" You can't show what isn't (necessarily) true. It's perfectly possible that $2x -1$ is negative. Consider 2 case: Case 1: $\sqrt M > w \ge 0$. Then $M > w^2$. But we must also consider $\sqrt M \ge 0 > w$. In a way this is easier. $w$ is irrelevant and $M > 0$. Combining the two you have $M >w^2$ and $w \ge 0$ OR $M \ge 0$ and $w < 0$. $\endgroup$
    – fleablood
    Oct 18, 2019 at 22:44

2 Answers 2

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Since $x^2+1>0$, the inequality

$$\sqrt{x^2+1}>2x-1$$

is always satisfied for $2x-1\le 0 \implies x\le \frac12$.

Then for $2x-1>0$ we can square to obtain

$$x^2+1>4x^2-4x+1 \iff3x^2-4x<0 \iff 0<x<\frac43$$

and we need to take the solutions compatible with the condition $2x-1>0$ that is $x>\frac12$.

Therefore solutions are

  • $x\le \frac12$
  • $\frac12<x<\frac43$

that is $x<\frac43$.

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    $\begingroup$ @BiliDebili The quantity under the square root is strictly positive, that is $x^2+1>0$. Then we can consider 2 cases: $2x-1\le 0$ and $2x-1>0$. In the first case $2x-1\le 0 \iff x\le \frac12$ the inequality always holds, since we have a positive term on the LHS and a negative term on the RHS. For the second case, since all quantities are positive we can square both sides and solve. $\endgroup$
    – user
    Oct 18, 2019 at 21:26
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It's obvious that for $x\leq0$ our inequality is true.

Let $x>0$.

Thus, we need to solve $$\sqrt{x^2+1}-x>x-1$$ or $$\frac{1}{\sqrt{x^2+1}+x}>x-1.$$ Now, consider two functions: $f(x)=\frac{1}{\sqrt{x^2+1}+x}$ and $g(x)=x-1$.

We see that $f$ decreases for $x>0$ and $g$ increases, which says that graphs of $f$ and $g$ intersect in one point maximum.

But $f\left(\frac{4}{3}\right)=g\left(\frac{4}{3}\right)$, which gives the answer: $$\left(-\infty,\frac{4}{3}\right).$$

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