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; unfortunately, I can't find them either.  Can someone tell me how to solve this or point me in the right direction with resources? 
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.  

$$\forall x,y \in \mathbb R,\quad xy \leq 2 \implies x \leq \sqrt 2 \lor y \leq \sqrt 2$$
 A: 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 $(A \implies B)$.
Contradiction
$A$ 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.
A: 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.
A: $$(A\Rightarrow B) \equiv (\lnot B\Rightarrow\lnot A)$$


*

*The statement is equivalent to its contraposition: if $x< 1$ and $y< 1$ and $z< 1$ then $x+y+z< 3$.

*Similarly, use contraposition.

A: 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 
$$M \leq \sqrt 2 $$ 
$$\square 2$$
