Proving basics of $(a+b)^2$ I need to prove this: 
Consider the following inequality:

$$a^2+ab+b^2 > 0$$

I know that $^2$ makes $a$ and $b$ positive numbers, so it always be $>0$, but i got stuck with the ab thing. I thought about $a^2+b^2+ab+ab=(a+b)^2,$ but no results... I also tried to think when the sum of $a^2+b^2$(not just a number) might bigger then $ab,$ but I'm having trouble using any of this to solve it.
Would be glad for any help!
p.s. sorry if i took it to the wrong section
SORRY to mention it! $b\neq0$ and $b,a$ are real numbers.
 A: *

*If $a>0$ and $b>0$ then obviously $a^2+b^2+ab>0$

*If $a<0$ and $b<0$ same as $1$ 

*If $a<0$ and $b>0$ assume $|a|>|b|$ thus $a^2-|a|b >0$ thus $a^2+ab+b^2>0$, if $|a|<|b|$ then $b^2-|a|b>0$ thus $a^2+b^2+ab>0$ 

*If $a>0$ and $b<0$ similar to $4$

A: $$a^2+ab+b^2=(a+\frac{1}{2}b)^2+\frac{3}{4}b^2$$
A: $$a^2+ab+b^2=\frac1{b^2}\left[\left(\frac ab\right)^2+\frac ab+1\right]$$
Now make the change $x=a/b$ and study the sign of the polynomial $x^2+x+1$.
A: Hint : $$ab\leq max\{a^2,b^2\}$$ then $(a+b)^2-ab>0$
A: a^2-ab+b^2>0
Can be written
a^2+b^2>ab
Case1: a is positive b is negative, which means the statement is true same goes for the reverse, since a^2+b^2 will always be positive and ab will always be negative.
Case2: a and b are the same sign and one is greater than the other.
aa + bb > ab, if a is greater than b then aa>ab, if b is greater than a then bb > ab
For example 3*3 > 3*2  and if b is larger 5*5>4*5, if they are different one must be greater than the other which means one of either a^2 or b^2 will be greater than ab, the highest possible outcome would be when a=b which is the next case. 
Case3: a and b have the same sign and are the same number
aa + bb > ab can be rewritten as nn + nn > n*n which can be written as 2n^2>n^2
A: Focusing on the middle term, you can see that the quantity $a^2+ab+b^2$ lies $\frac 34$ of the way from $(a-b)^2$ to $(a+b)^2$, which are both non-negative, and which cannot both be zero, since $b\ne 0$.
Therefore it must be strictly positive.
