<s>[INST] proof-check: $\frac{x+a}{x^2+1}$ always has $3$ inflection points I am asked to show that the function $f(x) = \frac{x+a}{x^2+1}$ always has $3$ inflection points for any $a \in \mathbb{R}$.
The second derivative of $f$ turns out to be 
$$f''(x) = \frac{2(x^3 + 3ax^2 - 3x - a)}{(x^2 + 1)^3}.$$
The discriminant of that third degree polynomial is 
$$108a^4 + 216a^2 + 108 > 0$$
so the polynomial always has $3$ real distinct roots. 
Now, a 3rd degree polynomial with 3 distinct roots changes sign at it's roots, so the whole $f''$ must change sign at the roots of the polynomial, so the $3$ roots of the polynomial are indeed inflection points.
Is this enough?
 [/INST] Also since $g(x)=x^3+3ax^2-3x-a \implies g'(x)=3x^2+ 6ax-3=0 \implies x_{1,2}=-a\pm \sqrt{a^2+1}$ has one minimum at $x=x_1$  maximum at $x=x_2$ as $g''(x_1)=6 \sqrt{1+a^2}>0$ and  $g''(x_2)=-6 \sqrt{1+a^2}.$  Further, $g'(x_1)=6 \sqrt{1+a^2},$ $g(x_1)=-2(1+a^2)[\sqrt{1+a^2)}-a]<0$ and $g(x_2)=2(1+a^2)[\sqrt{1+a^2)}-a]>0.$
So as mim is negative and max is positive  $g(x)=0$ has three distinct real roots.
Eventually $f(x)$ has three points of inflexion. 
This is what @Simon Zhou seems to suggest
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