IRL ABooklet

Solution 1. Observe that
$$\begin{aligned} \frac{a}{(2a + 3)^2} &= \frac{a}{4a^2 + 12a + 9} = \frac{a}{4(a - 1)^2 + 20a + 5} \\ &\leq \frac{a}{20a + 5} = \frac{1}{20} \cdot \frac{4a}{4a + 1} = \frac{1}{20} - \frac{1}{20} \cdot \frac{1}{4a + 1}. \end{aligned}$$

So, the LHS of the original inequality does not exceed
$$\frac{3}{20} - \frac{1}{20} \left( \frac{1}{4a+1} + \frac{1}{4b+1} + \frac{1}{4c+1} \right).$$
It now is sufficient to prove
$$\frac{1}{4a+1} + \frac{1}{4b+1} + \frac{1}{4c+1} \ge \frac{3}{5},$$
because then
$$\frac{a}{(2a+3)^2} + \frac{b}{(2b+3)^2} + \frac{c}{(2c+3)^2} \le \frac{3}{20} - \frac{1}{20} \cdot \frac{3}{5} = \frac{3}{25}.$$
By the AM-HM or the Cauchy-Schwarz inequality,
$$\frac{1}{4a+1} + \frac{1}{4b+1} + \frac{1}{4c+1} \ge \frac{9}{4a + 4b + 4c + 3}.$$
Using $a^2 + b^2 + c^2 = 3$, the QM-AM inequality implies
$$\frac{a+b+c}{3} \le \sqrt{\frac{a^2+b^2+c^2}{3}} = 1,$$
hence $4a + 4b + 4c + 3 \le 15$ and we obtain
$$\frac{1}{4a+1} + \frac{1}{4b+1} + \frac{1}{4c+1} \ge \frac{9}{15} = \frac{3}{5},$$
as required.

Solution 2. Note that the function $f(x) = -x/(2x+3)^2$ is convex on the interval $[0, 3]$, since $f''(x) = \frac{8(3-x)}{(2x+3)^4} \ge 0$ for such $x$. Hence, we can use Jensen's inequality, $(f(a) + f(b) + f(c))/3 \ge f((a+b+c)/3)$, and obtain
$$\frac{1}{3} \left( \frac{a}{(2a+3)^2} + \frac{b}{(2b+3)^2} + \frac{c}{(2c+3)^2} \right) \le \frac{t}{(2t+3)^2},$$
where $t = \frac{a+b+c}{3}$. The QM-AM inequality implies $t \le \sqrt{\frac{a^2+b^2+c^2}{3}} = 1$. Using this and $(2t+3)^2 \ge (2t+3)^2 - (2t-2)^2 = 20t + 5 = 5(4t+1)$, we get
$$\frac{t}{(2t+3)^2} \le \frac{t}{5(4t+1)} = \frac{1}{20} \left( 1 - \frac{1}{4t+1} \right) \le \frac{1}{20} \left( 1 - \frac{1}{5} \right) = \frac{1}{25}$$
which gives the desired inequality.