I was reading about the AM-GM inequality at brilliant.org.
In one of the examples, it proves that:
$(a^2+1)(b^2+1)\ge 4ab , ab\in\mathbb{R^+}$
But I thought applying the AM-GM inequality to a term with a constant is invalid since $(a^2+1)$ can be written as diffrent terms like $(a^2+\frac{1}{2}+\frac{1}{2})$ which would result in a diffrent answer.
If im wrong, can someone explain to me where my logic is flaud?
Yes, this will result in a different answer, but those are inequalities – there is no such thing as one best answer. You will get many different estimates this way, and all of them might be useful in an appropriate situation.
In your case, applying AM-GM to $a^2, \frac{1}{2}, \frac{1}{2}$ will yield
$$ a^2 + 1 \geqslant \frac{3}{\sqrt[3]{4}} a^{\frac{2}{3}}. $$
This is still true, and it is not always worse (or not always better) than $2a$ bound. The latter one is just more appropriate to the problem you are solving.
Looking at that a little deeper, the issue is: how big do you expect $a$ to be?
In your problem, equality $a^2+1 \geqslant 2a$ holds only if $a=1$, so this value of $a$ is the one we should be paying attention to. If, however, we expect $a$ to be around $\frac{1}{2} \sqrt{2}$, then $2a$ is approximately $\sqrt{2}$, and our strange-looking expression $\frac{3}{\sqrt[3]{4}} a^{\frac{2}{3}}$ turns out to be $\frac{3}{2}$ – so our choice of $a^2, \frac{1}{2}, \frac{1}{2}$ gave us a better result! (an optimal one, in fact)
This is obviously the effect of AM-GM being optimal when variables are equal to each other. If you try to solve a problem via AM-GM, paying attention to the cases when we have equality is usually worth doing.
One last example when artificially breaking down a constant might help: if we are interested in estimating $\sqrt[n]{x}$, then by using AM-GM for $x$ and $1,1,\ldots,1$ (with $n-1$ ones) we get
$$ \frac{x+n-1}{n} \geqslant \sqrt[n]{x}, $$
which, again, is a good estimate near $x=1$, and it might be hard to obtain without the many-ones trick.