### 2011 Baltic Way #4

Let $a,b,c,d$ be non-negative reals such that $a+b+c+d=4$. Prove the inequality \[\frac{a}{a^3+8}+\frac{b}{b^3+8}+\frac{c}{c^3+8}+\frac{d}{d^3+8}\le\frac{4}{9}\] By the AM-GM inequality for 3-variables, we have $$a^3+2=a^3+1+1 \ge 3\sqrt[3]{a^3 \cdot 1 \cdot 1 }=3a.$$ Thus, it is enough to show that $$\frac{a}{3a+6}+\frac{b}{3b+6}+\frac{c}{3c+6}+\frac{d}{3d+6} \le \frac{4}{9}.$$ Note that we can write the last inequality as $$\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}+\frac{1}{d+2} \le \frac{4}{3}. \space \space \space \space \space \space \space \space (*)$$ By HM-AM inequality, we can say that $$\frac{1}{4}(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}+\frac{1}{d+2}) \ge \frac{4}{(a+2)+(b+2)+(c+2)+(d+2)}$$ $$=\frac{4}{4+2+2+2+2}=\frac{1}{3}.$$ This implies the desired inequality, namely $(*)$. $\square$