Solve $S= (a+b)^2 -3ab \geqslant 1$ for $a>0$ and $b>0$

Question

I am trying to solve $S= (a+b)^2 -3ab \geqslant 1$ for $a>0$ and $b>0$.

After dividing both sides by $ab$, I get $a^2+b^2 \geqslant 2ab$, which means $(a-b)^2 \geqslant 0$ which is always true.

This should also mean that $S=1$ if $a=b$, which, however, does not always hold when I take specific numeric examples. Can anyone tell me why, or where I went wrong?

Answer 1

If $a,b$ are integers, then $$ S = a^2-ab+b^2 = {1\over 2}\Big((a-b)^2+a^2+b^2\Big)\geq {1\over 2}(0+1+1) = 1$$

If they are not, then the statment is not true, take $a=b$ and let $a\to 0$.