<s>[INST] If a and b are non-negative real numbers then demonstrate inequality If $a$ and $b$ are non-negative real numbers then show that
$$(3a +\frac{4}{a+1}+\frac{8}{\sqrt{2(1+b^2)}})\cdot(3b +\frac{4}{b+1}+\frac{8}{\sqrt{2(1+a^2)}})\geq81.$$
Inequality is less common. To demonstrate the usual instruments (means inequality, ...) did not help. Does anyone have any idea?
 [/INST] Well... Notice that by the AM-GM inequality: 
$$ 3a+\frac{4}{a+1}=\left(a+1\right)+\frac{2\left(a^2+1\right)}{a+1}+1 \ge 2\sqrt{2\left(1+a^2\right)}+1 $$
Then using the Cauchy–Bunyakovsky–Schwarz inequality, we get 
$$ LHS \ge \left(2\sqrt{2\left(1+a^2\right)}+1+\frac{8}{\sqrt{2\left(1+b^2\right)}}\right)\left(\frac{8}{\sqrt{2\left(1+a^2\right)}}+1+2\sqrt{2\left(1+b^2\right)}\right) \ge 81 $$
as desired. The equality occurs when $a=b=1$.
</s>