Need help with a proof

Prove for every two distinct real numbers a and b, either

(a+b/2) > a

or

(a+b/2) > b

If b <0 your inequality fails.

I haven't done math proofs in years.  I remember you should quote a rule or make a logical conversion or something for each step.

1. a and b are distinct real numbers
2. a + b/2 > a is true
3. We can simplify that to b/2 > 0
4. when b = 0, 0/2 > 2 is false

1. a and b are distinct real numbers
2. a + b/2 > b is true
3. We can simplify that to a > b/2 and then to 2a > b
4. when a = 1 and b = 2, 2 > 2 is false

>>3
Messed up on that first one, I did:
4. when b = 0, 0/2 > 0 is false

We wish to show that a+b/2 > a or a+b/2 > b
Let a be positive then either b is positive and the result is trivial or b is negative, but a is positive so the second result is trivial.
Now let a be negative... and that is also trivial
However consider the case where a=b=0, the result fails

But... if you turn either of the inequalities into a non-strict inequality, the result holds

Oh right it says they're distinct. so a!=b

QED

NEED PROFF OF GILGAMESH

>>3
That parenthetical is incorrect.

If (a+b)/2 > a is true, then a/2 + b/2 > a \implies b/2 > a/2 \implies b > a

If (a+b)/2 > b is true, then a/2 + b/2 > b \implies a/2 > b/2 \implies a > b

So, we have that the first inequality is equivalent to (b > a) and the second is equivalent to (a > b)

Now we need to prove that the real numbers are well-ordered.

Fuck.