<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] Call the function you're minimizing $f(a,b)$.  Note that the first three terms are homogeneous of order $2$ in $a$ and $b$, while the fourth is homogeneous of order $1$.
Thus $$f(ta, tb) = t^2 \left(a^2 + b^2 + \frac{a^2 b^2}{(a+b)^2}\right) - 2 t \frac{a^2+ab+b^2}{a+b} = t^2 g(a,b) + t h(a,b)$$
where $g \ge 0$, and this is minimized with respect to $t$ when $t = -h(a,b)/(2 g(a,b))$,
the minimum value being $-h(a,b)^2/(4 g(a,b))$, which simplifies to $-1$.
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