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

Question

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

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

Answer 2

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: $$\sum_{cyc}(\frac{1}{a^2+3}\leq\frac{3}{\left(\frac{\sum\limits_{cyc}a}{3}\right)^2+3}=\frac{3}{\left(\frac{1}{3}\right)^2+3}=\frac{27}{28}$$

Answer 3

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