Power inequalities

A counterexample:

$x=-1 \Longrightarrow -1<(-1)^2=1 \Longrightarrow (-1)^2=1\not < (-1)^3=-1$

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If $0\leq x\leq 1$ , then $x\geq x^2$ , so $x<0$ or $x>1$ . If $x>1$ , then $x<x^2<x^3$ , so for $x>1$ the statement is true. However we still have a possible case: $x<0$ . Note that $(-a)^2=a^2$ . So it doesn't matter what is the value of $x$ (under $0$ ), $x$ and $x^3$ are negativ numbers, and $x^2$ is a positive number. So in this case $x^2\not < x^3$ .

Therefore the statement is sometimes true, sometimes false.

First condition:

If $x = 5$ .

$\Rightarrow x^2 = 5^2 = 25$ which is greater than 5.

$\Rightarrow x^3 = 5^3 = 125$ which is greater than $5^2 = 25$ .

It satisfies the condition.

Second condition:

If $x = -5$

$\Rightarrow x^2 = (-5)^2 = 25$ which is greater than -5.

$\Rightarrow x^3 = (-5)^3 = -125$ which is smaller than 25.

It doesn't satisfies the condition.

Hence the answer is $\boxed{no}$