THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD

First solution. We notice the equality case $a=3$, $b=c=d=0$ and we try to mix the variables as follows: $$f(a,b,c,d)\ge f(a+b+c+d,0,0,0)=a+b+c+d-\frac{(a+b+c+d{)}^2}{3}=0,$$ where $$f(a,b,c,d)=\frac{a}{1+2b^3}+\frac{b}{1+2c^3}+\frac{c}{1+2d^3}+\frac{d}{1+2a^3}-\frac{a^2+b^2+c^2+d^2}{3}.$$ Then $$f(a,b,c,d)-a-b-c-d+\frac{(a+b+c+d{)}^2}{3}=$$ $$\frac{a}{1+2b^3}+\frac{b}{1+2c^3}+\frac{c}{1+2d^3}+\frac{d}{1+2a^3}-a-b-c-d+\frac{2(ab+bc+cd+da+ac+bd)}{3}\ge 0$$ $$\frac{a}{1+2b^3}+\frac{b}{1+2c^3}+\frac{c}{1+2d^3}+\frac{d}{1+2a^3}-a-b-c-d+\frac{2(ab+bc+cd+da)}{3}=$$ $$a\left(\frac{1}{1+2b^3}-1+\frac{2b}{3}\right)+b\left(\frac{1}{1+2c^3}-1+\frac{2c}{3}\right)+c\left(\frac{1}{1+2d^3}-1+\frac{2d}{3}\right)+$$ $$d\left(\frac{1}{1+2a^3}-1+\frac{2a}{3}\right)=$$ $$\frac{2ab(2b^3-3b^2+1)}{3(1+2b^3)}+\frac{2bc(2c^3-3c^2+1)}{3(1+2c^3)}+\frac{2cd(2d^3-3d^2+1)}{3(1+2d^3)}+\frac{2ad(2a^3-3a^2+1)}{3(1+2a^3)}=$$ $$\frac{2ab(2b+1)(b-1{)}^2}{3(1+2b^3)}+\frac{2bc(2c+1)(c-1{)}^2}{3(1+2c^3)}+\frac{2cd(2d+1)(d-1{)}^2}{3(1+2d^3)}+$$ $$\frac{2ad(2a+1)(a-1{)}^2}{3(1+2a^3)}\ge 0.$$ We have equality when $ac+bd=0$ and also $a=3$, $b=c=d=0$ or $a=2$, $b=1$, $c=d=0$ as well as for their cyclic permutations.

Second solution. (Ionuț Nicolae) We use the inequality $\frac{1}{1+2a^3}\ge 1-\frac{2a}{3}$ which, after computations, comes to $a(a-1{)}^2(2a+1)\ge 0$. The last inequality is obviously true and its equality cases are $a=0$ and $a=1$. We have $\frac{d}{1+2a^3}\ge \frac{d-2ad}{3}$, with equality if $d=0$ or $a\in \{0,1\}$. Then ${\sum }_{cycl}\frac{a}{1+2b^3}\ge {\sum }_{cycl}\left(a-\frac{2ab}{3}\right)=3-\frac{2}{3}{\sum }_{cycl}ab$. It remains to be proven that $a^2+b^2+c^2+d^2+2(ab+bc+cd+da)\le 9$, which follows from $a^2+b^2+c^2+d^2+2(ab+bc+cd+da+ac+bd)=9$. We have equality if $ac+bd=0$ and each non-zero variable is followed in cyclic order by one that is either $0$ or $1$. From $ac+bd=0$ we see that two variables need to be $0$. If another one is $0$ then the last one is $3$ and we have indeed equality; if the other two are positive, the second one needs to be $1$, hence the first one must be $2$.

Third solution. (given in the contest by Andrei Pantea) From AM-GM inequality we have $1+2x^3=1+x^3+x^3\ge 3x^2,\forall x>0$. We treat the following cases: 1. $a$, $b$, $c$, $d>0$; 2. $a$, $b$, $c>0$, $d=0$; 3. $a$, $b>0$, $c=d=0$; 4. $a$, $c>0$, $b=d=0$; 5. $a=b=c=0$, $d=3$. All the other cases follow by cyclic permutations. Case 1. The inequality is equivalent to $$\sum _{cycl}\left(a-\frac{a}{1+2b^3}\right)\le \frac{1}{3}\left((a+b+c+d{)}^2-\sum _{cycl}{a}^2\right).$$ But $$\sum _{cycl}\frac{2a{b}^3}{1+2b^3}\le \sum _{cycl}\frac{2a{b}^3}{3b^2}=\frac{2}{3}\sum _{cycl}ab<\frac{2}{3}\sum _{cycl}ab=\frac{1}{3}\left((a+b+c+d{)}^2-\sum _{cycl}{a}^2\right),$$ so in this case the inequality is satisfied without equality.