can someone give the proof to this:
a*a + ab + b*b > 0
for a =/= 0 , b=/= 0.
A square number is always positive, so you have a^2 + b^2 > 0
No matter what a and b are, one number will be larger, so either a > b or b > a. So if for example if a < b, then b^2 will always be larger than ab, as firstly a square number is always positive, and secondly b x b is larger than a x b when a < b. So this means that the a^2 or b^2 will always be higher than ab can possibly be, so the number will always be above 0, then adding another square number cannot decrease the number, so adding the other value^2 will not bring it below 0.
My proof is rather messy, but I think you get the idea.