cyclic rational inequalities $\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}\leq\frac{27}{28}$ when $a+b+c=1$ I've been practicing for high school olympiads and I see a lot of problems set up like this:
let $a,b,c>0$ and $a+b+c=1$. Show that
$$\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}\leq\frac{27}{28}$$
Any problem that involves cyclic inequalities like these always stump me. I know I'm supposed to use Cauchy-Schwarz or AM-GM at some point, but I can never get to a place where this might be useful. My first instinct is to get common denominators and hope stuff simplifies, but I can never get farther than that. For example, in this problem I did the following:
$$\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}$$
$$=\frac{(a^2+3)(b^2+3)+(b^2+3)(c^2+3)+(c^2+3)(a^2+3)}{(a^2+3)(b^2+3)(c^2+3)}$$
$$=\frac{a^2b^2+b^2c^2+c^2a^2+6(a^2+b^2+c^2)+27}{(a^2+3)(b^2+3)(c^2+3)}$$
but this is where I get stuck. I've tried using Cauchy-Schwarz on parts of this fraction to simplify it, but I can never get anything to work. How could you prove this inequality, and what are the important things to look out for in problems of this nature

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 $\, a + b + c = 1)$.
