Find minimum of $a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b}$ If $a,b$ are real numbers, find the minimum value of:
$$a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b}$$
This is what I did: I tried some values and I set $a=0$. Then, it becomes a quadratic of $b$:
$$b^2-2b$$
Here, the minimum is $-1$. So, I tried to prove that:
$$a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b}\ge -1$$
Using Wolfram, I found this is a square:
$$a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b} + 1 = \frac{(a^2+b^2+ab-a-b)^2}{(a+b)^2} $$
so it is positive.
My question is, can we prove this with more traditional and natural solution, maybe with Cauchy-Schwarz?

Let $x=a+b$ and $y=\frac{ab}{a+b}$. 
\begin{align}
 a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b} &=(a+b)^2-2ab+\left(\frac{ab}{(a+b)}\right)^2-\frac{2((a+b)^2-ab)}{a+b}
\\&= x^2-2xy+y^2-2(x-y)=\underset{\text{quadratic in } x-y}{\underbrace{(x-y)(x-y-2)}}\\
&\geq -1.
\end{align}
