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
 A: 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
$$\frac{27}{28}-\sum_{cyc}\frac{1}{a^2+3}=\sum_{cyc}\left(\frac{9}{28}-\frac{9}{x^2+27}\right)=\frac{9}{28}\sum_{cyc}\frac{x^2-1}{x^2+27}=$$
$$=\frac{9}{28}\sum_{cyc}\left(\frac{x^2-1}{x^2+27}-\frac{1}{14}(x-1)\right)=\frac{9}{392}\sum_{cyc}\frac{(x-1)^2(13-x)}{x^2+27}\geq0.$$
A: 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 $\, a + b + c = 1)$.
A: Another way.
Since $$\left(\frac{1}{x^2+3}\right)''=\frac{6(x^2-1)}{x^2+3)^2}<0$$ for $0<x<1$,  by Jensen for the concave function $f(x)=\frac{1}{x^2+3}$ we obtain:
$$\sum_{cyc}(\frac{1}{a^2+3}\leq\frac{3}{\left(\frac{\sum\limits_{cyc}a}{3}\right)^2+3}=\frac{3}{\left(\frac{1}{3}\right)^2+3}=\frac{27}{28}$$
