<s>[INST] A problem on inequality If real numbers $a >b >1$ satisfy the inequality
$$ ( ab+1)^{2} +( a+b)^{2} \leqslant 2( a+b)\left( a^{2} -ab+b^{2} +1\right) $$
What is the minimum possible value of $\frac{\sqrt{a-b}}{b-1}$?
My approach:
$$ ( a+b)\left( a^{2} -ab+b^{2} +1\right) =( a+b)\left( a^{2} -ab+b^{2}\right) +( a+b) =a^{3} +b^{3} +a+b $$
So, the question becomes $ ( ab+1)^{2} +( a+b)^{2} \leqslant 2\left( a^{3} +b^{3}\right) +2( a+b) $ 
 [/INST] The condition gives
$$2a^3-(b^2+1)a^2-2a(2b-1)+2b^3-b^2+2b-1\geq0$$ or
$$a^2(2a-b^2-1)-2a(2b-1)+(2b-1)(b^2+1)\geq0$$ or
$$(2a-b^2-1)(a^2-2b+1)\geq0$$ and since $$a^2-2b+1>a^2-2a+1=(a-1)^2>0,$$ we obtain $$2a\geq b^2+1$$ or
$$2(a-b)\geq(b-1)^2$$ or
$$\frac{\sqrt{a-b}}{b-1}\geq\frac{1}{\sqrt2}.$$
The equality occurs for $(a,b)=(2.5,2)$ for example, which says that $\frac{1}{\sqrt2}$ it's a minimal value.
</s>