Find the real solutions of the following equation
$$\frac{4x-1}{x^2-2x+2}+\frac{4x+7}{x^2+2x+2}=\frac{4(2x+3)}{x^2+2}.$$
I think the problem was given in a junior olympiad.
My best idea was to denote $x-1=a$, $x+1=b$ and after adding $2$ the LHS becomes
$$\frac{(a+2)^2}{a^2+1}+\frac{(b+2)^2}{b^2+1}$$
which looks like a Cauchy-Schwarz inequality. Unfortunately, I couldn't find a good form for the RHS.
I also tried $x^2-2x+2=a$, $x^2+2x+2=b$ which leads to a somewhat nice looking
$$\frac{b-1}{a}+\frac{7-a}{b}=\frac{4(b-a+6)}{a+b}$$
but the computations are not nice.
By plotting, there are four solutions among which are $\pm\sqrt{2}$.
Best Answer
By multiplying with a common denominator the equation is equivalent to $$ - 3x^4 + 8x^3 + 12x^2 - 16x - 12=0 $$ By writing it as $-(3x^2+ax+b)(x^2+cx+d)$ and comparing coefficients we see that this is just $$ (3x^2 - 8x - 6)(x^2 - 2)=0. $$ This gives four real solutions.