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I have a question relating to inequalities.

If $x$ is not less than $y$, can we definitely say that $x$ is greater than or equal to $y$? Is it necessary that $x$ exists? Compared to the condition $x$ is greater than or equal to $z$, which is stronger?

I am not asking about the existence of the inequalities themselves, but rather, their strength as a condition.

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    The answer depends on what are $x,y,z$. Are they real numbers? Are they complex numbers? Are they triangles? – 2017-01-21
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    Yes: if not $x < y$, then $x \ge y$. The negation of "less than" is "greater-or-equal". – 2017-01-21
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    The issue is not with "inequalities"... $x < y$ is simply a formula expressing a binary relation $R(x,y)$ that holds (or not) between "elements". If $n,m$ are e.g. natural numbers, we can always assert that $n < m$ : for some $n,m$ it is true, for other it is not. But it makes little sense to assert a relation between non-existent "objects". – 2017-01-21

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You can't say "if $x$ is not less than $y$" without implicitly assuming that you are talking about some particular $x$ and $y$. Then that statement is exactly the same as "(that particular) $x$ is greater than $y$".

(All assuming that you are talking about integers or rationals or reals, where "greater than" makes sense and you always have exactly one of $x < y$, $x =y$ or $x > y$.

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    Are there any situations when x does not necessrily exist if it is not less than or equal to y? Or does that make any logical sense? – 2017-01-21
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    I don't think that makes much logical sense as you say it. Of course it is true that there is no even prime number $x$ not less than or equat to $2$. But why would you say that this way? – 2017-01-21