Inequalities And Nothing Else

Relevant wiki: Polynomial Inequalities - Problem Solving - Medium

We have $6$ cases for this inequality:

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Case 1 : For $x > 1$ , we have

$x<x^2<x^3$

Example: $x=2,\;x^2=4,\;x^3=8\;,2<4<8$

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Case 2 : For $0<x<1$ , we have

$x^3<x^2<x$

Example: $x=0.1,\;x^2=0.01,\;x^3=0.001,\;0.001<0.01<0.1$

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Case 3 : For $x=0,1$ , we have

$x=x^2=x^3$

Proof: $x=0,\;x^2=0,\;x^3=0\quad\quad\quad x=1,\;x^2=1,\;x^3=1$

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Case 4 : For $x=-1$ , we have

$x=x^3<x^2$

Proof: $x=-1,\;x^2=1,\;x^3=-1,\;-1<1$

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Case 5 : For $-1<x<0$ , we have

$x<x^3<x^2$

Example: $x=-0.1,\;x^2=0.01,\;x^3=-0.001,\;-0.1<-0.001<0.01$

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Case 6 : For $x<-1$ , we have

$x^3<x<x^2$

Example: $x=-2,\;x^2=4,\;x^3=-8,\;-8<-2<4$

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Now, let's check back the given options.

$x<x^2<x^3$ is satisfied by Case 1

$x^3<x<x^2$ is satisfied by Case 6

$\boxed{x^2<x^3<x}$ is not satisfied by any of the cases above, therefore this inequality cannot be true

I shall prove that $x^2<x^3<x$ cannot be true for all real $x$ . Consider the first part $x^2<x^3$ . This implies that $x^3$ is positive which implies that $x>0$ . Now, consider the second part $x^3<x$ . We can hence deduce that $0<x<1$ . Then, since $0<x<1$ , $x^n<x^{n-1}$ for $n\in \mathbb{N}$ . Hence, there is a contradiction, and the proposition is proven.

I used process of elimination. For the first one, any real number greater than or equal to one works. For the last one, any negative number works. So the only one left is the middle statement since the other two are sometimes true.