is the sum of the square of two real numbers greater than or equal to twice the product of the two real numbers? I'm learning how to write proofs and cant seem to figure out how to do this one. Specifically Im interested if it is possible to prove the inequality by contradiction, contraposition, or by a direct proof; or if its possible at all.
 A: $$x^2+y^2-2xy = x(x-y) + y(y-x) $$
$$= (x-y)(x-y)=(x-y)^2 \ge 0$$
Thus
$$x^2 +y^2 \ge 2xy,\, \forall x,y \in \mathbb R$$
A: Whenever you are asked to figure out if $A$ is greater than $B $ there are two things you should consider if you don't know what else to do:
1) look at $\frac AB $ and see if it is greater or less than $1$ (but be VERY careful about observing if $B $ is positive or negative or zero.
2) Look at $A-B $ and see if it is greater or less than $0$.
Doing 1) we get $\frac {a^2+b^2}{2ab} $.   (But only if $a $ and both are both non-zero!  And we don't know if $ab $ is positive or negative.) Maybe or maybe not the geometric vs arithmetic mean inequality will jump out at you. ($\frac {x+y}2\ge \sqrt{xy} $ for positive $x,y $)  .... I'll forgive you if it doesn't.  It didn't jump out at me either.  And maybe you haven't learned it yet.
But I'd be remiss if I didn't point out that $\frac {a^2+b^2}2\ge \sqrt {|a^2b^2|}=|ab|\ge ab $ so $\frac {a^2+b^2}{2|ab|} \ge 1$ and $a^2+b^2\ge 2|ab|\ge 2ab $.  I'd be especially remiss as I'm the one who told you to look in the first place.
But that was the hard way and I wasn't actually trying to teach you to see that.  And again, if you haven't learned the AM-GM inequality yet, none of that will make sense.
But what I was really getting at was doing 2).  
We get $a^2+b^2-2ab $.  Now that should really look like something very familiar.  It should scream  $a^2+b^2-2ab=a^2-2ab+b^2=(a-b)^2$.
Now what were we doing?  Oh, yes.  We were trying to see if $a^2+b^2-2ab=(a-b)^2$ was greater or less than  $0$.
Well.... it's a square.  Squares are always greater or equal to $0$.  So $a^2+b^2\ge 2ab $.  And $(a-b)^2=0$ only if $a=b $.  So $a^2+b^2 >2ab $ unless $a=b $ (in which case they are equal...obviously).
A: You START with a true statement, and then show that it implies what you want to prove. 
For all real numbers $x$ and $y$, $(x-y)^2 \ge 0$.
Therefore, $x^2 - 2xy + y^2 \ge 0$.
And, finally, therefore, for all real numbers $x$ and $y$,  $x^2 + y^2 \ge 2xy$.
A: Translated into a statement you have
$$ a^2 + b^2 \ge 2ab $$
(subtract $2ab$ from both sides) iff 
$$ a^2 - 2ab + b^2 \ge 0 $$
using the fact that $a^2-2ab+b^2 = (a-b)^2$, iff 
$$ (a-b)^2 \ge 0 $$
Since $a$ and $b$ are real, then $a-b$ is real (reals are closed under subtraction). And any real number squared must be greater than or equal to $0$. So this statement is true for any $a$ and $b$.
A: Try to write out what do you want to show in mathmetical term.
$$\forall a, b \in \mathbb{R}, a^2+b^2 \geq 2ab$$
While a direct proof is the obvious way for me, I shall purposely write the proof in contradiction format for exposure purpose:
suppose it is not true, then $\exists a,b \in \mathbb{R}, a^2+b^2 < 2ab$ 
 which implies that $(a-b)^2 <0$, which is a contradiction.
