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For example, if you have $x > 6$ and $y > 4$, then obviously $x + y > 10$. If you have $x > 6$ and $y < 4$, then the second inequality can be changed to $-y > -4$. Now adding $x > 6$ gives $x-y > 2$, which is obviously true. Now suppose you have $x + y > 6$ and $x - y > 4$. Adding the inequalities, the $y$ terms cancel and you end up with $2x > 10$ or $x > 5$. It is not intuitively obvious to me that this holds true, but I could not come up with a counter-example. I can see that you can't subtract inequalities, but is it always okay to add them?

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Consider any two inequalities $a < b$ and $c < d$, where each of $a$, $b$, $c$, and $d$ is real. Since $a < b$, it must be the case that (colloquially speaking) shifting each of them the same distance up or down the real line won’t disrupt things. In other words, $a + c < b + c$. (Note that this last bit wasn’t a proof, but merely an attempt to help you hone your intuition.)

But by the exact same kind of reasoning, it follows from $c < d$ that $b + c < b + d$. And now, all we need is the transitivity of less-than: Since

$$a + c < b + c < b + d,$$

we conclude that $a + c < b + d$.

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  • $\begingroup$ Looks like we were typing the same thing at the same time, down to the choice of variables! Great minds think alike. :-) $\endgroup$
    – JonathanZ
    Sep 1, 2023 at 22:03
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I'll assume you know that you can add the same thing to both sides of an inequality, no? That is, if you know that $$ x \lt y$$ then $$ x +z \lt y+z$$ is valid.

Now, your question is: If I know that $$a \lt b$$ and $$c \lt d$$ does this imply that $$ a+c \lt b+d \text{ ?}$$

It does, and here's how to prove it. Add $c$ to both sides of your first inequality to get $$a+c \lt b+c$$ and then add $b$ to both sides of your second inequality to get $$ b+c \lt b+d$$ and then just use transitivity ($r \lt s$ and $s \lt t$ imply $r \lt t$) on those last two inequalities to get the inequality you wanted.

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  • $\begingroup$ Then it is perfectly okay to add inequalities. In particular, the original solution I gave in this question is perfectly reasonable, though maybe not intuitively obvious. math.stackexchange.com/questions/4729228/… I was also thinking that you could go from equations to inequalities by saying that x > y is equivalent to saying that x = y + p for some positive number p. $\endgroup$ Sep 1, 2023 at 23:43
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Say you have two bags of cookies, $a$ and $b$. A friendly baker comes by and offers to trade with you: you will give the baker your bag $a$ and in return you will get a larger bag $c$ which contains more cookies. That is, $a<c$. You like cookies, so you agree.

Now the friendly baker offers the same deal for bag $b$: you will give up $b$ and in return you get a bigger bag $d$ that again contains more cookies. That is, $c<d$. You make that trade and the baker leaves.

Later you open the bags and count up your cookies. You have $c+d$ now, and you had $a+b$ before. Is it possible that you might not have more cookies than before you made the trades?

No, it's obviously impossible.

But that's what it would mean if $a<c$ and $b<d$ but not $a+b<c+d$ too. So that is also impossble.

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