<s>[INST] 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?
 [/INST] To prove that:
$$a^2+b^2+\frac{a^2b^2}{(a+b)^2}+1 \ge \frac{2(a^2+ab+b^2)}{a+b}$$
we can use AM-GM:
$$
\begin{aligned}
a^2+b^2+\frac{a^2b^2}{(a+b)^2}+1 &\geq 2\sqrt{a^2+b^2+\frac{a^2b^2}{(a+b)^2}}\\
&= 2\sqrt{\frac{(a^2+b^2)^2+2(a^2+b^2)ab+a^2b^2}{(a+b)^2}}\\
&= 2\sqrt{\frac{(a^2+ab+b^2)^2}{(a+b)^2}}\\
&=2\left|\frac{a^2+ab+b^2}{a+b}\right|\\
&\geq \frac{2(a^2+ab+b^2)}{a+b}
\end{aligned}
$$
Equality occurs when $a^2+ab+b^2=a+b$.
Later edit: As a matter of fact, we have 
$$a^2+b^2+\frac{a^2b^2}{(a+b)^2}=\frac{(a^2+ab+b^2)^2}{(a+b)^2}$$
therefore, if you substitute $x=\dfrac{a^2+ab+b^2}{a+b}$, the question is rephrased as minimize the function $f(x)=x^2-2x$.
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