<s>[INST] Given $abc=1$ and $0< c \leq b \leq1\leq a$, prove that $8(a+b+c)^2\le9(1+a^2)(1+b^2)(1+c^2)$ I can't make progress with proving this inequality.
I have tried opening the brackets and using $abc=1$ in order to obtain the following:
$$a^2+b^2+ \frac 1{a^2b^2}+18+9a^2b^2+\frac 9{a^2}+\frac9{b^2}\ge 16 \left (ab+\frac 1a+\frac 1b \right)$$
 [/INST] Since the right side of your inequality does not depend on substitution $a\rightarrow-a$, 
it's enough to prove it for positive variables.
We'll prove that your inequality is true for all positives $a$, $b$ and $c$ such that $abc=1$.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, our inequality is equivalent to 
$$8\cdot9u^2w^2\leq18w^4+9(9u^2-6v^2)w^2+9(9v^4-6uw^3)$$ or
$f(v^2)\geq0$, where
$$f(v^2)=9v^4-6v^2w^2-6uw^3+u^2w^2+2w^4$$
Since $f'(v^2)=18v^2-6w^2>0$, it remains to prove our inequality for a minimal value of $v^2$,
which happens for equality case of two variables.
Let $b=a$ and $c=\frac{1}{a^2}$.
We obtain $(a-1)^2(9a^6+18a^5+13a^4+8a^3+21a^2+2a+1)\geq0$.
Done!
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