IRL_ABooklet_2023

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

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 $f^{\prime \prime }(x)=\frac{8(3-x)}{(2x+3{)}^4}\ge 0$ 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.