Question 170
Find the minimal possible value of $\frac{\sqrt{a-b}}{b-1}$ over all real numbers $a \gt b \gt 1$ satisfying
$$
(ab+1)^2 + (a+b)^2 \le 2(a+b)(a^2 - ab + b^2 + 1).
$$
$$
(ab+1)^2 + (a+b)^2 \le 2(a+b)(a^2 - ab + b^2 + 1).
$$
MathNet 2017-part-06 — 176 problems (EN–VI) Question 170
Math Olympiad Vault · MathNet 2017-part-06 — 176 problems (EN–VI)