<s>[INST] Why is my proof for “if $0 \leqslant x \leqslant 2$, then $-x^3 + 4x + 1 > 0$” is false? 
$$\text{if $0 \leqslant x \leqslant 2$, then $-x^3 + 4x + 1 > 0$}$$

$$x(4-x^2)>-1$$
$$x>\dfrac{1}{x^2-4}$$ if the last statement is smaller than $x$ then it is also smaller than $2$ because of first statement ($0 \leqslant x \leqslant 2$):
$$\dfrac{1}{x^2-4}<2$$
$$1<2x^2-8$$
$$4.5<x^2$$
$$-2.121<x<2.121$$
Which contradicts the first statement because $x$ cannot be greater than $2$. My proof is wrong but why? (I took that question from Math for Computer Science book p. 12)
 [/INST] You made two mistakes and recovered:
$$\dfrac{1}{x^2-4}<2 \not\Rightarrow 1<2(x^2-4) \ \ (\text{it must be} \ 1\color{red}>2(x^2-4), \text{because} \ x^2-4\le 0)\\
4.5<x^2 \not\Rightarrow -2.121<x<2.121 \ (\text{it must be} \ 4.5\color{red}>x^2 \Rightarrow -2.121<x<2.121)$$
Alternatively, you can prove it as follows:
$$-x^3 + 4x + 1 > 0 \iff x^3-4x-1<0 \iff x(x-2)(x+2)<1 \iff \\
x(x-2)(x+2)\le 0<1 \ \ \text{for} \  0\le x\le 2 \ \text{(why?)}$$
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