<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] We begin by observing that if $a=b=c=1$, then the two expressions are equal to each other. 
Expand everything. Use the equation $abc=1$ to get rid of compound terms. Gather terms with each variable separately. Use the preceding observation to factor each nominator.
$$\begin{align} & 9(1+a^2)(1+b^2)(1+c^2)-8(a+b+c)^2 \\ &=9\bigl(1+a^2+b^2+c^2+a^2b^2+b^2c^2+c^2a^2+a^2b^2c^2\bigr)-8\bigl(a^2+b^2+b^2+2ab+2bc+2ca\bigr) \\ &= 9\Bigl(1+a^2+b^2+c^2+{1\over c^2}+{1\over a^2}+{1\over b^2}+1\Bigr)-8\Bigl(a^2+b^2+c^2-{2\over c}-{2\over a}-{2\over b}\Bigr) \\ & ={a^4+6a^2-16a+9 \over a^2}+{b^4+6b^2-16b+9 \over b^2}+{c^4+6c^2-16c+9 \over c^2} \\ & ={(a-1)^2(a^2+2a+9) \over a^2}+{(b-1)^2(b^2+2b+9) \over b^2}+{(c-1)^2(c^2+2c+9) \over c^2}\end{align}$$
It is now obvious that this last expression is always $\geq 0$, and that it is $=0$ only in the case when $a=b=c=1$. 
</s>