# Why doesn't my solution give the right answer

I'm trying to solve this inequality:

$$1-\frac{2}{\left| x\right|}\leqslant \frac{23}{x^{2}}$$

Let's imagine we already know the answer: $$x ∈ [-1-2\,\sqrt{6}; 0)⋃(0; 1+2\,\sqrt{6}]$$

Now I am going to present my solution (wrong solution):

$$1-\frac{2}{\left| x\right|}\leqslant \frac{23}{x^{2}}$$ ; $$1 = \frac{x^{2}}{x^{2}}$$ $$(1)$$

$$\frac{23-x^{2}}{x^{2}}+\frac{2}{\left| x\right|}\geqslant 0$$ $$(2)$$

And now I will multiply the both sides by adjoint expression:

$$\left(\frac{23-x^{2}}{x^{2}}+\frac{2}{\left| x\right|}\right)\cdot \left(\frac{23-x^{2}}{x^{2}}-\frac{2}{\left| x\right|}\right)\geqslant 0$$ $$(3)$$

Using $$\left(a-b\right)\,\left(a+b\right)=a^{2}-b^{2}$$ and the fact that $$\left| x\right|^{2}=x^{2}$$ I will proceed with my solution.

I stopped on the 3rd step because my solution is already incorrect on it. I am curious why.

• The problem is what if that adjoint expresseinon is negative. Then the inequality sign reverses. Commented Nov 10, 2022 at 14:34
• You know that $a+b>0$, but not $a-b>0$. So you cannot deduce $(3)$. Commented Nov 10, 2022 at 14:34

As you noticed $$|x|^2 =x^2$$ so why not use that. Let $$t=|x|>0$$, since $$x=0$$ is not a solution. Then we have to solve $$1-{2\over t}\le {23\over t^2}$$ and thus $$t^2-2t+1\le 24\implies 0 and the rest is easy...