<s>[INST] 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
 [/INST] 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
$$\frac{27}{28}-\sum_{cyc}\frac{1}{a^2+3}=\sum_{cyc}\left(\frac{9}{28}-\frac{9}{x^2+27}\right)=\frac{9}{28}\sum_{cyc}\frac{x^2-1}{x^2+27}=$$
$$=\frac{9}{28}\sum_{cyc}\left(\frac{x^2-1}{x^2+27}-\frac{1}{14}(x-1)\right)=\frac{9}{392}\sum_{cyc}\frac{(x-1)^2(13-x)}{x^2+27}\geq0.$$
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