67th NMO Selection Tests for JBMO

Notice that $a^3+b^3+1\ge 3ab$ and $b^3+1+1\ge 3b$ to obtain that $\frac{1}{a^3+2b^3+6}\le \frac{1}{3ab+3b+3}$, hence it is sufficient to show that $\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\le 1$.

To this end, observe that $\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}=\frac{1}{ab+b+1}+\frac{ab}{a{b}^{2}c+abc+ab}+\frac{b}{abc+ab+b}\stackrel{abc\ge 1}{\le }\frac{1}{ab+b+1}+\frac{ab}{b+1+ab}+\frac{b}{1+ab+b}=\frac{1+ab+b}{ab+b+1}=1$.

Alternative Solution:

Subtract $1/6$ from each of the left hand-side summands and write successively ${\sum }_{cyc}\left(\frac{1}{a^3+2b^3+6}-\frac{1}{6}\right)\le \frac{1}{3}-\frac{1}{2}$, then ${\sum }_{cyc}\frac{-a^3-2b^3}{6(a^3+2b^3+6)}\le -\frac{1}{6}$ or further ${\sum }_{cyc}\frac{a^3+2b^3}{a^3+2b^3+6}\ge 1$.

To this end, notice that $$\begin{array}{rl}\sum _{cyc}\frac{a^3}{a^3+2b^3+6}& =\sum _{cyc}\frac{a^4}{a^4+2a{b}^3+6a}\\ & \stackrel{CBS}{\ge }\frac{(a^2+b^2+c^2{)}^2}{a^4+b^4+c^4+2(a{b}^3+b{c}^3+c{a}^3)+6(a+b+c)}\stackrel{(1)}{\ge }\frac{1}{3}.\end{array}$$

The inequality (1) rewrites $3(a^4+b^4+c^4)+6(a^2{b}^2+b^2{c}^2+c^2{a}^2)\ge a^4+b^4+c^4+2(a{b}^3+b{c}^3+c{a}^3)+6(a+b+c)$, and follows from $2(a^4+b^4+c^4)\ge 2(a{b}^3+b{c}^3+c{a}^3)$ and $6(a^2{b}^2+b^2{c}^2+c^2{a}^2)\ge 6(ab\cdot bc+bc\cdot ca+ca\cdot ab)=6abc(a+b+c)\ge 6(a+b+c)$.

On the other hand, $$\begin{array}{rl}\sum _{cyc}\frac{b^3}{a^3+2b^3+6}& =\sum _{cyc}\frac{b^4}{b{a}^3+2b^4+6b}\\ & \stackrel{CBS}{\ge }\frac{(a^2+b^2+c^2{)}^2}{2(a^4+b^4+c^4)+(b{a}^3+c{b}^3+a{c}^3)+6(a+b+c)}\\ & \stackrel{(2)}{\ge }\frac{1}{3}.\end{array}$$

The inequality (2) rewrites $3(a^4+b^4+c^4)+6(a^2{b}^2+b^2{c}^2+c^2{a}^2)\ge 2(a^4+b^4+c^4)+(b{a}^3+c{b}^3+a{c}^3)+6(a+b+c)$, and follows from $a^4+b^4+c^4\ge b{a}^3+c{b}^3+a{c}^3$ and $6(a^2{b}^2+b^2{c}^2+c^2{a}^2)\ge 6(ab\cdot bc+bc\cdot ca+ca\cdot ab)=6abc(a+b+c)\ge 6(a+b+c)$.

Alternative Solution:

As $a^3+2b^3=a^3+b^3+b^3\ge 3a{b}^2$, it suffices to prove that $\frac{1}{a{b}^2+2}+\frac{1}{b{c}^2+2}+\frac{1}{c{a}^2+2}\le 1$. Rewrite the inequality as $-\frac{a{b}^2}{2(a{b}^2+2)}-\frac{b{c}^2}{2(b{c}^2+2)}-\frac{c{a}^2}{2(c{a}^2+2)}\le 1-\frac{3}{2}$, or, equivalently, $$\frac{a{b}^2}{a{b}^2+2}+\frac{b{c}^2}{b{c}^2+2}+\frac{c{a}^2}{c{a}^2+2}\ge 1.$$ Recall that $abc\ge 1$, and write $\frac{a{b}^2}{a{b}^2+2abc}+\frac{b{c}^2}{b{c}^2+2abc}+\frac{c{a}^2}{c{a}^2+2abc}\ge 1$ to observe that it is enough to show that $\frac{b}{b+2c}+\frac{c}{c+2a}+\frac{a}{a+2b}\ge 1$. Indeed, $\frac{b}{b+2c}+\frac{c}{c+2a}+\frac{a}{a+2b}=\frac{b^2}{b^2+2bc}+\frac{c^2}{c^2+2ca}+\frac{a^2}{a^2+2ab}\stackrel{CBS}{\ge }\frac{(a+b+c{)}^2}{b^2+2bc+c^2+2ca+a^2+2ab}=1$, which concludes the proof.