Inequality – Prove $\frac{1}{x^3+2}+\frac{1}{y^3+2}+\frac{1}{z^3+2}\leq 1$

Question

question

Let $x,y,z$ be positive numbers such that $xyz\geq 1$. Prove that $\frac{1}{x^3+2}+\frac{1}{y^3+2}+\frac{1}{z^3+2}\leq 1$.

my idea

if we look closely we realise that we have to prove that

$\frac{1}{x^3+2}+\frac{1}{y^3+2}+\frac{1}{z^3+2}\leq xyz$

The first thing I thought of is the inequlity of means or CBS

A thing i demonstrated by Am-Gm is

$x^3+2=3x$

I think thismight help us. I dont know what to do forward. Hope one of you can help me! Thank you

Answer 1

Let $x=ka,$ $y=kb$ and $z=kc,$ where $k>0$ and $abc=1$.

Thus, $$k^3abc\geq abc,$$which gives $k\geq1$ and $$\sum_{cyc}\frac{1}{x^3+2}=\sum_{cyc}\frac{1}{k^3a^3+2}\leq\sum_{cyc}\frac{1}{a^3+2}$$and it's enough to prove that $$\sum_{cyc}\frac{1}{a^3+2}\leq1$$ or $$\sum_{cyc}\frac{1}{a^3+2abc}\leq\frac{1}{abc}$$ or $$\sum_{cyc}\frac{bc}{a^2+2bc}\leq1,$$ which is true by C-S: $$1-\sum_{cyc}\frac{bc}{a^2+2bc}=-\frac{1}{2}-\sum_{cyc}\left(\frac{bc}{a^2+2bc}-\frac{1}{2}\right)=$$ $$=-\frac{1}{2}+\sum_{cyc}\frac{a^2}{2(a^2+2bc)}\geq-\frac{1}{2}+\frac{(a+b+c)^2}{2\sum\limits_{cyc}(a^2+2bc)}=0.$$ I used C-S in the Engel's form:

Let $a_i$ real numbers and $b_i>0$. Prove that: $$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+...+\frac{a_n^2}{b_n}\geq\frac{(a_1+a_2+...+a_n)^2}{b_1+b_2+...+b_n}.$$