<s>[INST] Proving $\sum_{cyc}\sqrt[3]{\frac{1}{a}+\frac{2}{bc}+a+2b+c}\leq\frac{6}{abc}$ for positive values such that $ab+bc+ca=3$ This problem is from my teacher.

Know that $a,b,c>0$, $ab+bc+ca=3$.
Prove that: $$\sum_{cyc}\sqrt[3]{\frac{1}{a}+\frac{2}{bc}+a+2b+c}\leq\frac{6}{abc}$$

I tried  to change the number '$6$' into $2(ab+bc+ca)$.
Then get
$$\sum_{cyc}\sqrt[3]{\frac{1}{a}+\frac{2}{bc}+a+2b+c}\leq\frac{2ab+2bc+2ca}{abc}=\frac{2}{c}+\frac{2}{b}+\frac{2}{a}$$
but i have no idea what to do next.
 [/INST] By holder: $$\sum_{cyc}\sqrt[3]{\frac{1}{a}+\frac{2}{bc}+a+3b+c}\le \sqrt[3]{\sum_{cyc}({\frac{1}{a}+\frac{2}{bc}+a+3b+c})(1+1+1) \cdot (1+1+1)}$$ It hence suffices to prove $$\sqrt[3]{\sum_{cyc}({\frac{1}{a}+\frac{2}{bc}+a+3b+c})(1+1+1)(1+1+1)}\le \frac{6}{abc}$$ $$\sum_{cyc}({\frac{1}{a}+\frac{2}{bc}+a+3b+c})\le \frac{24}{a^3b^3c^3}\tag 1$$ Now let $p=a+b+c,q=ab+bc+ca=3,r=abc$ then we rewrite (1) as $$\frac{q}{r}+\frac{2p}{r}+5p\le \frac{24}{r^3}$$ $$\iff qr^2+2pr^2+5pr^3\le 24$$ Now as $q^2\ge 3pr \iff p\le \frac{3}{r}$  and $q=3$
it suffices to prove $$3r^2+6r+15r^2\le 24$$ which is true as $r\le 1$(by AM-GM)

Note in your original question you had $...a+2b+c$  i think its a typo and should be $a+3b+c$
all the same since $$\sum_{cyc}\sqrt[3]{({\frac{1}{a}+\frac{2}{bc}+a+2b+c})}\le \sqrt[3]{\sum_{cyc}({\frac{1}{a}+\frac{2}{bc}+a+3b+c})}$$ the same proof applies here here so we are done in either case
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