I copied the example out but I am not interested in the example per say, except for the change in inequality that takes place. I underlined it in red.

The Example

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My Question

What I understand: (for RHS)

$f(x) - g(x) < \frac{f(p)}{2}$

$\implies g(x) >f(x) - \frac{f(p)}{2}$

so how did get "$\geq$"?

  • 5
    $\begingroup$ Surely $A > B$ implies $A \geq B.$ I don't see any particular reason why it's not "$>$" in the second underlined inequality, but writing $\geq$ doesn't spoil the proof. $\endgroup$
    – David K
    Jun 27, 2022 at 2:02
  • $\begingroup$ @DavidK, thank you for your reply. But shouldn't $A > B$ mean that $A$ never actually $= B$, but $A \geq B$ implies that $A$ can $= B$. Allowing this feels like room for some misplay, or am I not fully understanding what "$A > B$" really means? $\endgroup$
    – Reuben
    Jun 27, 2022 at 2:08
  • 2
    $\begingroup$ I see there is a full answer now, so I don't need to explain. But for what it's worth, I think $>$ would have been preferable to $\geq$. It just happens that the proof still holds up, but it introduces an extraneous "or $g(x)=f(x)-\frac{f(p)}{2}$" which we know has no effect since $g(x)\neq f(x)-\frac{f(p)}{2}$, and the proof would be cleaner if this extraneous "or" weren't there. $\endgroup$
    – David K
    Jun 27, 2022 at 2:31
  • $\begingroup$ If "Adam is an anteater" is true, then "Adam is an anteater or a zebra" is also true. "Adam is an anteater" is a stronger statement because it is telling you that Adam is most assuredly and anteater and the second is only telling you that "Adam is either an anteater or a zebra". But a weaker statement is still a true statement... After all if it were not true that $A \ge B$ it what have to be true that $A < B$ and that is WRONG. So $A \ge B$ is true. $\endgroup$
    – fleablood
    Jun 27, 2022 at 2:42
  • 2
    $\begingroup$ Actually, I bet that was just a type-setting error that never was caught. I bet the manuscript did say $g(x) > f(x) - \frac {f(p)}2$. (But what is written is not technically wrong). $\endgroup$
    – fleablood
    Jun 27, 2022 at 2:44

2 Answers 2


Consider two numbers $x$ and $y$. Then $x\geq y$ implies either $x=y$ OR $x>y$ holds.
So, if $x>y$ is true then so is $x\geq y$ since $x\geq y$ requires at least one of $x=y$ OR $x>y$ to be true. Think of it in terms of (mathematical) logic.
Thus, if the statement $g(x)>f(x)-\dfrac{f(p)}{2}$ is true, then this implies that the statement $g(x)\geq f(x)-\dfrac{f(p)}{2}$ is also automatically true.

  • $\begingroup$ For example, it is true to say that $1 \geq 2$, even though it would be a strange thing to say $\endgroup$
    – Prince M
    Jun 27, 2022 at 6:27
  • 2
    $\begingroup$ @PrinceM Do you mean $1\leq 2$? $\endgroup$
    – QC_QAOA
    Jun 27, 2022 at 6:42

As already mentioned in previously submitted answer, logically you can say $x > y$ can also be written as $x ≥ y$ because at least one of the cases is valid. According to this $5 ≥ 2$ also taken as correct. But I think it's better to include equal sign if there is a possibility for it to happen because it is a special case.
Let's consider an integer $n$, can you say $n > 5$ is same as $n ≥ 5$? I suggest you can't because in the first case minimum of $n$ is $6$, but in the second case minimum of $n$ is $5$, therefore the implications are totally different.
Let's consider another situation,
Does it imply $a ≥ c$ given that $a ≥ b > c$ or $a ≥ c$ given that $a ≥ b ≥ c$?


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