Find minimum of $a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b}$

Question

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?

Answer 1

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 $f(x)=x^2-2x$.

Answer 2

If you see this as a map $\mathbb{R}^2 \to \mathbb{R}$, it is differentiable on $\mathbb{R}^2 \backslash \{a=-b\}$. You can calculate the critical points and the hessian to determine if it is a minimum. This only works to find a local minimum, so you should also look at the behaviour around the line $a=-b$

Answer 3

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 $-1$.

Answer 4

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}