$a$ and $b$ are both Real numbers (different than $0$)
I saw this problem in the math Olympiads of my home country
What I’ve tried so far:
Factoring $a+b+2 \sqrt{ab}$ to $(\sqrt{a} +\sqrt{b})^{2}$
Expanding the whole equation
And thank you for your help
Divide the equation by $a$, let $r=b/a$. This gives $$1+r+2\sqrt r = 2(r-1)$$ set $x=\sqrt r>0$ and we see a quadratic $$1+x^2 + 2x = 2(x^2-1) \iff 0 = x^2 - 2x -3 = (x-3)(x+1)$$ the positive solution is $x=3$, so $\frac ba = r = x^2 = 9$.
The conditions are that either $(a<0\land b<0)$ or $(a>0\land b>0)$. If the former, then $$\begin{cases}a<0\\ b<0\\ a+b+2\sqrt{ab}=2(b-a)\end{cases}\Leftrightarrow\begin{cases}a<0\\ b<0\\-(\sqrt{-a}-\sqrt{-b})^2=2(\sqrt{-a}-\sqrt{-b})(\sqrt{-a}+\sqrt{-b})\end{cases}\Leftrightarrow\\\Leftrightarrow\begin{cases}a<0\\ b<0\\(\sqrt{-a}-\sqrt{-b})(3\sqrt{-a}+\sqrt{-b})=0\end{cases}\Leftrightarrow \begin{cases}a<0\\ b<0\\ a=b\end{cases}\lor\begin{cases}a<0\\ b<0\\ \sqrt{-b}=-3\sqrt{-a}\end{cases}\Leftrightarrow\begin{cases}a<0\\ a=b\end{cases}$$
If they are both positive, $$\begin{cases}a>0\\ b>0\\ a+b+2\sqrt{ab}=2(b-a)\end{cases}\Leftrightarrow\begin{cases}a>0\\ b>0\\ (\sqrt a+\sqrt b)^2-2(\sqrt b-\sqrt a)(\sqrt a+\sqrt b)=0\end{cases}\Leftrightarrow\\\Leftrightarrow\begin{cases}a>0\\ b>0\\ (\sqrt a+\sqrt b)(3\sqrt a-\sqrt b)=0\end{cases}\Leftrightarrow\begin{cases}a>0\\ b>0\\ \sqrt a=-\sqrt b\lor 3\sqrt a=\sqrt b\end{cases}\Leftrightarrow\\\Leftrightarrow \begin{cases}a>0\\ b>0\\ b=9a\end{cases}$$
So, for non-zero real numbers the identity holds if and only if either $b=a<0$ or $b=9a>0$.
We have $(b-a) = (\sqrt{b} - \sqrt{a})(\sqrt{b} + \sqrt{a})$
So with your step we get:
\begin{align*}&(\sqrt{b} + \sqrt{a})^2 = 2(\sqrt{b} - \sqrt{a})(\sqrt{b} + \sqrt{a}) \\ \iff &(\sqrt{b} + \sqrt{a}) = 2(\sqrt{b} - \sqrt{a}) \\ \iff & 9 = \frac{b}{a} \end{align*}
Dividing $a+b+2\sqrt{ab}=2(b-a)$ by $a$ gives $$1+b/a+2\sqrt{b/a}=2b/a-2.$$ Hence, $3-b/a+2\sqrt{b/a}=0$. This is a quadratic equation in $\sqrt{b/a}$.
Divide throughout by $a$.
So,
$\begin{align}1+ \frac{b}{a} + 2\sqrt{\frac ba} = 2 \frac ba -2 \Rightarrow \frac ba - 2\sqrt{\frac ba} = 3 \end{align}$
$\Rightarrow\sqrt{\frac ba} = 3 $ or $-1$.
As $-1$ is not possible, $\sqrt{\frac ba} = 3$
or $\frac ba = 9$
Notice that the left side of the equation is actually equal to a square $$(\sqrt{a} + \sqrt{b})^2 = 2(b-a)$$
Now apply the usual $c^2 - d^2 = (c+d)(c-d)$
$$(\sqrt{a} + \sqrt{b})^2 = 2(\sqrt{a} + \sqrt{b})(-\sqrt{a} + \sqrt{b})$$
and finally
$$\sqrt{a} + \sqrt{b}= 2\sqrt{b} - 2\sqrt{a}$$
from which
$$\frac{\sqrt{b}}{\sqrt{a}} = 3$$
And hence your claim that $\frac{b}{a} = 9$.
$$\begin{align} (-a) + (-b) -2\sqrt{(-a)(-b)} &= 2[(-b)-(-a)]\ \left(\sqrt{-b} - \sqrt{-a}\right)^2 &= 2\left(\sqrt{-b}+\sqrt{-a}\right)\left(\sqrt{-b} - \sqrt{-a}\right) \end{align}$$
Then $\left(\sqrt{-b} - \sqrt{-a}\right)$ may be zero and cannot be simply rejected.
– peterwhy Jan 25 '21 at 12:57$$\begin{align} \frac{2\sqrt{ab}}{a} = \frac{2\sqrt{ab}}{-|a|} = \frac{2\sqrt{ab}}{-\sqrt{a^2}} = -2\sqrt{\frac{b}{a}} \end{align}$$
– peterwhy Jan 25 '21 at 13:06