Unbalanced Symmetric Inequality

Suppose $a \ge b \ge c$ . Let $a=s+t,b=s-t$ with $0 \le t \le s-t$ (because $a \ge b \ge c$ ).

Let $f(t)=\frac{c(s-t)}{3+(s+t)^2}+\frac{c(s+t)}{3+(s-t)^2}+\frac{s^2-t^2}{3+c^2}$ with $0 \le t \le s-t$ .

We have $f(t)=\frac{-c}{3+(s+t)^2}-\frac{2c(s^2-t^2)}{[3+(s+t)^2]^2}+\frac{c}{3+(s-t)^2}+\frac{2c(s^2-t^2)}{[3+(s-t)^2]^2}+\frac{2t}{3+c^2}$ .

Let $u=3+(s+t)^2,v=3+(s-t)^2$ . We have:

$f'(c)=\frac{4cst}{uv}+\frac{8cst(s^2-t^2)(u+v)}{u^2v^2}-\frac{2t}{3+c^2}$ .

We also have: $a+b=2s \implies 1 \le s \le 1.5$ (because $a \ge b \ge c$ ).

$\implies cs=(3-2s)s \le \frac{(3-2s+s)^2}{4}=\frac{(3-s)^2}{4} \le \frac{(3-1)^2}{4} = 1$ .

We have $u=3+(s+t)^2 \ge 3+(1+0)^2 = 4$ and $v=3+(s-t)^2 \ge 3+[s-(s-c)]^2= 3+c^2$ (because $0 \le c \le s-c$ ).

$\implies \frac{4cst}{uv} \le \frac{1}{3+c^2}(*)$

According to Cauchi-Schwarz inequality, we have:

$u^2v^2=[3+(s+t)^2]^2[3-(s-t)^2] \le 4^4(s^2-t^2)$ (1)

$8cs(u+v)(3+c^2)=4.4cs(3+s^2+t^2)(3+c^2) \le 4(\frac{4cs+3+s^2+t^2+3+c^2}{3})^3$ .

We also have

$4cs+3+s^2+t^2+3+c^2 \le 4(3-2s)s+6+s^2+(3s-3)^2+(3-2s)^2=12+6(s-1)(s-2) \le 12$ (because $c=3-2s, t^2 \le s-c=3s-3$ ).

Therefore $8cs(u+v)(3+c^2) \le 4(\frac{12}{3})^3=4^4 (2)$

(1),(2) $\implies \frac{8cst(s^2-t^2)(u+v)}{u^2v^2}=\frac{8cs(3+c^2)(u+v)}{3+c^2}.\frac{s^2-t^2}{u^2v^2} \le \frac{s^2-t^2}{4^4(s^2-t^2)}.\frac{4^4}{3+c^2}=\frac{1}{3+c^2}(**)$ .

$(*),(**)$ we have $f'(t) \le \frac{2}{3+c^2}-\frac{2t}{3+c^2}=(1-t)\frac{2}{3+c^2}\le 0$ with $t \in [0,s-c]$ .

$\implies f(t)<0$ with $t \in [0,s-c]$ $\implies f(t)<f(0)$ with $t \in [0,s-c]$ .

Let $g(s)=f(0)=\frac{2cs}{3+s^2}+\frac{s^2}{3+c^2}=\frac{3s(3-2s)}{3+s^2}+\frac{s^2}{3+(3-2s)^2}$ with $s \in [0,1.5]$ . We have:

$g'(s)=\frac{(6-8s)(3+s^2)-2s(6s-4s^2)}{(3+s^2)^2}+\frac{(2s[3+(3-2s)^2]-s^2(8s-12)}{[3+(3-2s)^2]^2}=\frac{108(s^2-3s+4)(s-1)^2(-s^2-3s+6)}{(3+s^2)^2[3+(3-2s)^2]^2}$ .

Take a look at $s^2-3s+4=(s-1.5)^2+1.75 > 0$ with $s \in [0,1.5]$ and $-s^2-3s+6=(\frac{\sqrt{33}-3}{2}-s)(\frac{\sqrt{33}+3}{2}+s)$

We can easily see that $Max(g(s))=\frac{11\sqrt{33}-45}{24}$ when $s=\frac{\sqrt{33}-3}{2}$ .

In conclusion the maximum value of the expression above is $\frac{11\sqrt{33}-45}{24}$ when $a=b=\frac{\sqrt{33}-3}{2}$ and $c=6-\sqrt{33}$ .

We will submit the answer as $A+B+C+D=11+33+45+24=\boxed{113}$

And Phew, we are done! I don't think this problem should be judged as "easy", even for a level 5 algebra