ComplexPhi wrote: Let $a$ , $b$ , $c$ be positive real numbers such that $abc=1$ . Show that : \[\frac{1}{a^3+bc}+\frac{1}{b^3+ca}+\frac{1}{c^3+ab} \leq \frac{ \left (ab+bc+ca \right )^2 }{6}\] hello, it can be proved by BW. Sonnhard.

Dr Sonnhard Graubner wrote: ComplexPhi wrote: Let $a$ , $b$ , $c$ be positive real numbers such that $abc=1$ . Show that : \[\frac{1}{a^3+bc}+\frac{1}{b^3+ca}+\frac{1}{c^3+ab} \leq \frac{ \left (ab+bc+ca \right )^2 }{6}\] hello, it can be proved by BW. Sonnhard. Can you post your full solution please?

By AM-GM, $a^3+bc\ge 2a$. We want to show that \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le \frac{(ab+bc+ca)^2}{3}\iff ab+bc+ca\le \frac{(ab+bc+ca)^2}{3}\iff \]\[\iff 3\le ab+bc+ca,\]which is true by AM-GM.

ComplexPhi wrote: Let $a$ , $b$ , $c$ be positive real numbers such that $abc=1$ . Show that : \[\frac{1}{a^3+bc}+\frac{1}{b^3+ca}+\frac{1}{c^3+ab} \leq \frac{ \left (ab+bc+ca \right )^2 }{6}\] The following inequality is also true. Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that: \[\frac{1}{3a^3+bc}+\frac{1}{3b^3+ca}+\frac{1}{3c^3+ab} \leq \frac{ \left (ab+bc+ca \right )^2 }{12}\]

Using the fact that $\frac{1}{a^3+bc} \leq \frac{1}{2a} = \frac{bc}{2}$ and $\sum ab =\sum \frac{1}{a}$ , we only have to prove that $3(\sum \frac{1}{a} ) \leq ( \sum \frac{1}{a} )^{2} <==> \sum \frac{1}{a} \geq 3 $ wich follows directly from am-gm since $abc=1$

By AM-GM we have: $a^3+bc\ge2\sqrt{a^3bc}=2\sqrt{a^2*abc}=2\sqrt{a^2*1}=2\sqrt{a^2}=2a$ $b^3+ca\ge2\sqrt{b^3ca}=2\sqrt{b^2*abc}=2\sqrt{b^2*1}=2\sqrt{b^2}=2b$ $c^3+ab\ge2\sqrt{c^3ab}=2\sqrt{c^2*abc}=2\sqrt{c^2*1}=2\sqrt{c^2}=2c$ $=>$ $\frac{1}{a^3+bc}\leq\frac{1}{2a}$ $\frac{1}{b^3+ca}\leq\frac{1}{2b}$ $\frac{1}{c^3+ab}\leq\frac{1}{2c}$ $\frac{1}{a^3+bc}+\frac{1}{b^3+ca}+\frac{1}{c^3+ab}\leq\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}$ $\frac{1}{a^3+bc}+\frac{1}{b^3+ca}+\frac{1}{c^3+ab}\leq\frac{1}{2}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ $\frac{1}{a^3+bc}+\frac{1}{b^3+ca}+\frac{1}{c^3+ab}\leq\frac{1}{2}(\frac{ab+bc+ac}{abc})$ $\frac{1}{a^3+bc}+\frac{1}{b^3+ca}+\frac{1}{c^3+ab}\leq\frac{ab+bc+ac}{2abc}$ $\frac{1}{a^3+bc}+\frac{1}{b^3+ca}+\frac{1}{c^3+ab}\leq\frac{ab+bc+ac}{2abc}*1$ $\frac{1}{a^3+bc}+\frac{1}{b^3+ca}+\frac{1}{c^3+ab}\leq\frac{ab+bc+ac}{2abc}*\frac{ab+bc+ac}{ab+bc+ac}$ $\frac{1}{a^3+bc}+\frac{1}{b^3+ca}+\frac{1}{c^3+ab}\leq\frac{(ab+bc+ac)^2}{2abc(ab+bc+ac)}$ $\frac{1}{a^3+bc}+\frac{1}{b^3+ca}+\frac{1}{c^3+ab}\leq\frac{(ab+bc+ac)^2}{2*1(ab+bc+ac)}$ $\frac{1}{a^3+bc}+\frac{1}{b^3+ca}+\frac{1}{c^3+ab}\leq\frac{(ab+bc+ac)^2}{2(ab+bc+ac)} ...(1)$ Now again from AM-GM we have: $ab+bc+ac\ge3\sqrt[3]{ab*bc*ac}$ $ab+bc+ac\ge3\sqrt[3]{a^2b^2c^2}$ $ab+bc+ac\ge3\sqrt[3]{(abc)^2}$ $ab+bc+ac\ge3\sqrt[3]{1}$ $ab+bc+ac\ge3*1$ $ab+bc+ac\ge3$ $=>$ $\frac{1}{ab+bc+ac}\leq\frac{1}{3} ...(2)$ Combining $(1)$ with $(2)$ we get $\frac{1}{a^3+bc}+\frac{1}{b^3+ca}+\frac{1}{c^3+ab}\leq\frac{(ab+bc+ac)^2}{2*3}$ $\frac{1}{a^3+bc}+\frac{1}{b^3+ca}+\frac{1}{c^3+ab}\leq\frac{(ab+bc+ac)^2}{6}$

@above your solution is equivalent to #4, #6, #7. Just a suggestion for the next time you can use the sum notation, also you can try to not write symmetric expressions as it is very clear they are true, and what their sum gives you.

We have, $$\frac{1}{a^3+bc}+\frac{1}{b^3+ca}+\frac{1}{c^3+ab}=\frac{bc}{a^3bc+(bc)^2}+\frac{ca}{ab^3c+(ca)^2}+\frac{ab}{abc^3+(ab)^2}=\frac{bc}{a^2+(bc)^2}+\frac{ca}{b^2+(ca)^2}+\frac{ab}{c^2+(ab)^2}$$By $AM-GM$, we have $$a^2+(bc)^2\ge 2\sqrt{a^2 \cdot (bc)^2}=2$$$$b^2+(ca)^2\ge 2\sqrt{b^2 \cdot (ca)^2}=2$$$$c^2+(ab)^2\ge 2\sqrt{c^2 \cdot (ab)^2}=2$$Thus, $$\frac{bc}{a^2+(bc)^2}+\frac{ca}{b^2+(ca)^2}+\frac{ab}{c^2+(ab)^2} \le \frac{ab+bc+ca}{2}$$By $AM-GM$ again, we have $$ab+bc+ca\ge 3\sqrt[3]{ab\cdot bc \cdot ca}=3 \Leftrightarrow \frac{(ab+bc+ca)^2}{6}\ge \frac{ab+bc+ca}{2}$$Therefore, $$\frac{1}{a^3+bc}+\frac{1}{b^3+ca}+\frac{1}{c^3+ab}=\frac{bc}{a^2+(bc)^2}+\frac{ca}{b^2+(ca)^2}+\frac{ab}{c^2+(ab)^2} \le \frac{ab+bc+ca}{2} \le \frac{(ab+bc+ca)^2}{6}$$The equality holds when $a=b=c=1$.