Question 21
Let $a$, $b$, and $c$ be positive real numbers with sum $3$. Prove that:
$$
\frac{a b}{a b+a+b}+\frac{b c}{b c+b+c}+\frac{c a}{c a+c+a}+\frac{1}{9}\left(\frac{(a-b)^{2}}{a b+a+b}+\frac{(b-c)^{2}}{b c+b+c}+\frac{(c-a)^{2}}{c a+c+a}\right) \leqslant 1
$$
$$
\frac{a b}{a b+a+b}+\frac{b c}{b c+b+c}+\frac{c a}{c a+c+a}+\frac{1}{9}\left(\frac{(a-b)^{2}}{a b+a+b}+\frac{(b-c)^{2}}{b c+b+c}+\frac{(c-a)^{2}}{c a+c+a}\right) \leqslant 1
$$