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How can one solve for $x$ the inequality $(a \pm b)^2 \le (c \pm x)^2$? If squares were not there, one could easily break $(a + b) \le (c + x)$ and $(a - b) \le (c - x)$, and then solve as usual. But the square seem to make it complicated.

User101
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    What do you mean by the $\pm$'s? Is this a collection of four different inequalities? Two different inequalities? Do you want them to hold simultaneously, or do you want to solve the different inequalities separately? – Milten Jul 20 '22 at 10:26
  • Yeah, I want to treat them as four different inequalities. – User101 Jul 20 '22 at 10:31
  • $$(a+b)^2≤(c+x)^2$$ Tells you that

    $$a+b≤c+x \text{ or } -(a+b)≥c+x$$

    Same goes for the three other equations

     – Barbaud Julien Jul 20 '22 at 10:42
  • Ok, if $x^2 \le y^2$, then $x\le y$ is clear to me. How do you conclude the second? – User101 Jul 20 '22 at 10:51
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    It should not be clear to you, because it is not correct actually. If $-x>y$, you also have $x^2<y^2$ – Barbaud Julien Jul 20 '22 at 10:55
  • @BarbaudJulien, I think your last statement is not true. If $x=-3$, $y=2$, then not only $-x>y$ is true but also $x^2 >y^2$ is also true. – User101 Jul 20 '22 at 12:43
  • use $x^2 >y^2$ then $(x-y)(x+y)\ge0$ – emil agazade Jul 20 '22 at 14:23

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