Question 121
Suppose that $a$, $b$, $c \gt 0$ and $a^2 + b^2 + c^2 = 3$. Prove that
$$
\frac{a}{(2a + 3)^2} + \frac{b}{(2b + 3)^2} + \frac{c}{(2c + 3)^2} \le \frac{3}{25}.
$$
$$
\frac{a}{(2a + 3)^2} + \frac{b}{(2b + 3)^2} + \frac{c}{(2c + 3)^2} \le \frac{3}{25}.
$$