The tangent line trick

Finding the line. $f\left(\frac13\right)=\frac3{10}$ and $f'(a)=\frac{1-a^2}{(a^2+1)^2}$, so $f'\left(\frac13\right)=\frac{18}{25}$. The tangent is
$$\ell(a)=\frac3{10}+\frac{18}{25}\left(a-\frac13\right)=\frac{18}{25}a+\frac3{50}.$$

The one-variable inequality. For $a>0$ we claim $\frac a{a^2+1}\le\frac{18a}{25}+\frac3{50}$. Multiplying by $50(a^2+1)>0$ it reads $50a\le(36a+3)(a^2+1)$, i.e.
$$36a^3+3a^2-14a+3\ge0.$$
This factors as $(3a-1)^2(4a+3)\ge0$, true for $a>0$.

Summing. Add the three inequalities and use $a+b+c=1$:
$$\sum\frac a{a^2+1}\le\frac{18}{25}(a+b+c)+\frac9{50}=\frac9{10}.$$

Remark. The tangent line trick works when the constraint is linear and the function is locally below its tangent.

Lesson. Constraint $a+b+c=\text{const}$ and a sum $f(a)+f(b)+f(c)$: guess the equality case, compute the tangent at that point, and check the one-variable inequality.