If a and b are real non zero numbers such that a^2 + b^2 = 1, is it true that a + b > 1

any guidance would be greatly appreciated, I've found it holds on a tonne of data I'm working with and I was trying to show it to be true/false, but I seem to be running in circles :S

Cheers,

David

a > 0, b > 0, 2ab > 0

a^2 + b^2 = 1
a^2 + b^2 + 2ab > 1
(a + b)^2 > 1

Since both a + b and 1 are > 0, then just take square root of both sides:
a + b > 1

Yes, it is true

if a = -1/sqrt(2) = b
then a^2 + b^2 = 1/2 + 1/2 = 1
but a + b = -1/sqrt(2) -1/sqrt(2) = -2/sqrt(2)
if a = -1/sqrt(2) and b = 1/sqrt(2)
the a^2 + b^2 = 1/2 + 1/2) = 1
bur a + b = -1/sqrt(2) + 1/sqrt(2) = 0
So, there's two exceptions within the restrictions you've set.

Now, if you'd said that a and b are real, with a and b> 0....

Try to find out by using Pythagoras.
We know then c=1
and a^2+b^2=1

If you draw this on a piece of paper, you can see that you can follow the hypotenuse straight over or go the longer way along the katets a and b. Therfore a+b>1. It is a longer way.

How old are you?
Wow.
(-½√2)² = ?