Prove if sum of 3 numbers $\geq$ 3 then at least one them must be $\geq$1

I am helping out a friend who can't seem to get these proofs and he enlisted my help thinking I know this. Unfortunately, I have tried wrapping my head around it and I can't find a solution to these problems.

Can someone tell me how to solve this or point me in the right direction with resources? Thanks!

Question 1:

 Prove that for all real numbers x, y, and z, if x + y + z greater than or equal
to 3, then either x greater than or equal to 1 or y greater than or equal to 1
or z greater than or equal to 1.


$$\forall x,y,z\in \mathbb R, \quad x+y+z \geq 3 \implies x \geq 1 \lor y \geq 1 \lor z\geq1$$

Question 2:

Prove that for all real numbers x and y, if xy less than or equal to 2, then
either x less than or equal to square root of 2 or y less than or equal to
square root of 2.


$$\forall x,y \in \mathbb R,\quad xy \leq 2 \implies x \leq \sqrt 2 \lor y \leq \sqrt 2$$

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Here are three ways you can choose to prove conditionals (if-then statements). There are other ways but these are most common.

Direct Proof

If $A$, then $B$.

$(A \implies B)$

Assume $A$ is true, then show that when $A$ is true that $B$ must also be true. This seems like it should be the simplest way, but that is not always the case.

Contrapositive

If not $B$, then not $A$.

$(\neg B \implies \neg A)$

This is when you assume that $B$ is false, then show that when $B$ is false then $A$ must also be false. This works because $(\neg B \implies \neg A)$ is logically equivalent to $(A \implies B)$.

$A$ and not $B$.

$(A \land \neg B)$

This is when you assume that $A$ is true and that $B$ is false. Then go on with attempting to prove that assumption just to arrive at a contradiction, or an absurdity in the proof. By showing $(A \land \neg B)$ is a contradiction you have done enough to prove that $(A \implies B)$ so you can end your proof there.

I know this is all formal symbolic logic but it's quite useful in math proofs. Hope that helps.

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Proof 1.

By contradiction, x<1 and y<1 and z<1 implies x+y+z <3, so this assumption is false. Thus we have x>=1 or y>=1 or z>=1.

Proof 2 can be done in a similar way.

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Thanks @ZhiyongWang. I'll try to do #2 and post my response here. –  AAA Mar 3 '13 at 0:53
I have a feeling I have it wrong but here it is: IF X <2 and y <2 then that implies that xy greater than or equal to 2. Thus we have xy greater than or equal to 2? –  AAA Mar 3 '13 at 1:07
@AAA That's not right. If $x<2$ and $y<2$ how do you get $xy\ge 2$? I edited my answer with a possible reasoning you can use, which is the one both me and Zhiyong have in mind. –  Git Gud Mar 3 '13 at 1:18

$$(A\Rightarrow B) \equiv (\lnot B\Rightarrow\lnot A)$$

1. The statement is equivalent to its contraposition: if $x< 1$ and $y< 1$ and $z< 1$ then $x+y+z< 3$.
2. Similarly, use contraposition.
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Thanks @Berci. I'll try to do #2 and post my response here. –  AAA Mar 3 '13 at 0:55
@GitGud: yes, absolutely right. Was too late.. –  Berci Mar 3 '13 at 10:48
@Berci Comment deleted. –  Git Gud Mar 3 '13 at 15:34

Proof of part 1: Let $M$ be the largest value of $x,y,z$ then at most there are 3 different values : $M,M-a,M-b$ where $a,b \geq 0$.

Now writing $x+y+z \geq 3$ in terms of $M$ we have $$M+(M-a)+(M-b) \geq 3$$

$$3M-a-b \geq 3$$ $$3M \geq 3 +a + b$$ $$M \geq 1 + \frac {a}{3} + \frac{b}{3} \implies M \geq 1$$ Since $a,b \geq 0 \implies \frac {a}{3} + \frac{b}{3}\geq0$ Since M was the maximum of $x,y,z$ then at least one of them $\geq 1$ $$\square1$$

Proof of part 2: For part 2 we have : Let $M$ be the lowest value of $x,y$ then at most there are 2 different values $M,M\alpha$ where $\alpha \geq 1$

Now writing $xy \leq 2$ in terms of $M$ we have $$M(M\alpha)\leq 2$$
$$M^2\alpha\leq 2$$
$$M\leq \sqrt \frac{2}{\alpha}$$
But since $\alpha \geq 1 \implies \frac{2}{\alpha} \leq 2$ we have $$M \leq \sqrt \frac{2}{\alpha}\leq \sqrt 2$$
or just $$M \leq \sqrt 2$$ $$\square 2$$

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