cyclic rational inequalities $\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}\leq\frac{27}{28}$ when $a+b+c=1$

Tangent Line method helps.

Indeed, let $a=\frac{x}{3},$ $b=\frac{y}{3}$ and $c=\frac{z}{3}.$

Thus, $x+y+z=3$ and $$\frac{27}{28}-\sum_{cyc}\frac{1}{a^2+3}=\sum_{cyc}\left(\frac{9}{28}-\frac{9}{x^2+27}\right)=\frac{9}{28}\sum_{cyc}\frac{x^2-1}{x^2+27}=$$ $$=\frac{9}{28}\sum_{cyc}\left(\frac{x^2-1}{x^2+27}-\frac{1}{14}(x-1)\right)=\frac{9}{392}\sum_{cyc}\frac{(x-1)^2(13-x)}{x^2+27}\geq0.$$

We need to prove $\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}\leq\frac{27}{28}$ for $a, b, c \gt 0, a + b + c = 1$

Using tangent line method,

We consider the equation of the tangent line to $f(x) = \frac{1}{3+x^2}$ at $x = \frac{1}{3}$. Point is $(\frac{1}{3}, \frac{9}{28})$

$f'(x) = -\frac{2x}{(3+x^2)^2} = -\frac{27}{392}$

So equation of tangent line $y = -\frac{27}{392} x+ c$

Given the point on the line, $y = -\frac{27}{392} x + \frac{135}{392}$

We claim that $f(x) = \frac{1}{3+x^2} \leq -\frac{27}{392} x + \frac{135}{392}$ ...(i)

it is equivalent of saying $\frac{(135-27x)(3+x^2)}{392} \geq 1$ for $0 \lt x \leq 1$

which is true and equality occurs for $x = \frac{1}{3}$.

Now we know at $x = \frac{1}{3}, f(x) = \frac{9}{28}$

So, $\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}\leq\frac{27}{28}$

EDIT: you could do it from (i) as follows too

$f(a) = \frac{1}{3+a^2} \leq -\frac{27}{392} a + \frac{135}{392}$ (same for $b$ and $c$)

$\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3} \leq -\frac{27}{392} (a + b + c) + \frac{3 \times 135}{392} \leq \frac{27}{28} \, ($as $\, a + b + c = 1)$.