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I know this probably seems like an easy problem, but I generally struggle with inequalities, so I was looking to see if someone could verify that the following proof of mine is correct. This was an even problem in my book, so no answers are provided. Please note that the variable $x$ is in the set of all real numbers.

Proof by contrapositive

Note: $x^3 - x > 0 \iff x(x+1)(x-1) > 0$

1) $x \leq -1$

2) $x + 1 \leq 0$

3) $x - 1 \leq -2$

4) $x(x+1) \geq 0$

5) $x(x+1)(x-1) \leq 0$

6) Therefore, if $x^3 - x > 0$, then $x > -1$

K Split X
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Aleksandr Hovhannisyan
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    Seems ok, but why not simply prove $x^3 \leq x$? Seems much easier. – Demophilus Jun 13 '17 at 22:49
  • Wow, that sure simplifies things :D Thank you! – Aleksandr Hovhannisyan Jun 13 '17 at 22:59
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    Nice. Also, <----> can be expressed as dollarsign /iff dollarsign – K Split X Jun 13 '17 at 23:45
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    Do not make a series of assertions with no explanation of how they are related. If you did that in everyday speech nobody would understand you. Write (i) $x\leq -1\implies x+1\leq 0.$...(ii). By (i), $x\leq -1\implies x(x+1)\geq 0$...(iii). $x\leq -1\implies x-1\leq -2\implies x-1\leq 0$....(iv). By (ii) and (iii), $x\leq -1\implies x^3-x=[x(x+1)][x-1] \leq 0.$... The extra brackets in (iv) are for clarity: (ii) says something about $x(x+1)$ and (iii) says something about $x-1,$ so in (iv) we separate them by extra brackets for easier reading. – DanielWainfleet Jun 14 '17 at 00:46
  • @DanielWainfleet Thank you for the advice! – Aleksandr Hovhannisyan Jun 14 '17 at 01:07
  • @KSplitX Gotcha, thanks for the edit :) – Aleksandr Hovhannisyan Jun 14 '17 at 01:10

2 Answers2

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Your proof is correct, but it can be streamlined.

If $x\le -1$, then $x^2\ge-x$, so $x^3\le-x^2$. But $-x^2\le x$ and therefore $$ x^3\le x $$ that is, $x^3-x\le0$.

egreg
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Using the general inequality laws $a\le b\implies {1\over a}\ge{1\over b}$ when $a$ and $b$ have the same sign, $a\ge b\implies ac\ge bc$ if $c\ge0$, and $x^2\ge0$ for all $x$, we have

$$x\le-1\implies{1\over x}\ge-1\ge x\implies x={x^2\over x}\ge x\cdot x^2=x^3\implies x^3-x\le0$$

Barry Cipra
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