Details on figuring out if my logic about this proof shows that I am done Proving that if a and b are positive real numbers then $a + b$ is greater than or equal to $\sqrt{ab}$
I put: $a > 0$ and $b > 0$, using direct proof
Proof: 
$(a+b)^2\geq ab$
$a^2 + 2ab + b^2\geq ab$
$a^2 + ab + b^2\geq 0$ which is true if $a+b\geq\sqrt{ab}$
 A: I believe you may have a correct argument in mind. However, the flow of the logic is not clear, partly because no words are used. The first line is
$$(a+b)^2\ge ab.$$
Are you asserting that this is true, at least for all positive $a,b$? If so, there ought to be justification.
The next line is
$$a^2+2ab+b^2\ge ab.$$
This certainly follows from the inequality of the previous line, but that inequality has not been proved. Perhaps you intend to say that the two inequalities are equivalent, one holds if and only if the other holds.
The next inequality $a^2+ab+b^2\ge 0$ is clearly equivalent to the previous one. You undoubtedly saw that $a^2+ab+b^2\ge 0$  obviously holds for positive $a$ and $b$. But you did not say so.
Now, after a couple of lines of working backwards, you probably consider the inequality $(a+b)^2\ge ab$ proved. You say this is true if $a+b\ge \sqrt{ab}$. Probably you intend the opposite implication, that from $(a+b)^2\ge ab$ we can conclude that $a+b\ge \sqrt{ab}$.
The logic should be made much clearer. One should also minimize or eliminate the working backwards, or at least make it very clear when that is being done.
For writing a quick proof that uses the ideas, note that if $a$ and $b$ are positive, then $2ab\ge ab$ and therefore $(a+b)^2\ge ab$. Now taking the square root of both sides yields the desired result.
