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Algebra Difficulty 3.9 AMC 10/12 Prove it Turkey

Find the minimal possible value of a−bb−1\frac{\sqrt{a-b}}{b-1} over all real numbers a>b>1a > b > 1 satisfying
(ab+1)2+(a+b)2≤2(a+b)(a2−ab+b2+1). (ab+1)^2 + (a+b)^2 \le 2(a+b)(a^2 - ab + b^2 + 1).

Solution

Using the given equation, we get
0≥(ab+1)2+(a+b)2−2(a+b)(a2−ab+b2+1)=a2b2+4ab+1−2a3−2b3+a2+b2−2a−2b=(a2−2b+1)(b2−2a+1) \begin{aligned} 0 &\ge (ab+1)^2 + (a+b)^2 - 2(a+b)(a^2 - ab + b^2 + 1) \\ &= a^2b^2 + 4ab + 1 - 2a^3 - 2b^3 + a^2 + b^2 - 2a - 2b \\ &= (a^2 - 2b + 1)(b^2 - 2a + 1) \end{aligned}
Since a>ba > b, we get a2−2b+1>b2−2b+1=(b−1)2≥0a^2 - 2b + 1 > b^2 - 2b + 1 = (b-1)^2 \ge 0 and b2−2a+1≤0b^2 - 2a + 1 \le 0. Therefore, we have (b−1)2≤2(a−b)(b-1)^2 \le 2(a-b) or
a−bb−1≥12 \frac{\sqrt{a-b}}{b-1} \ge \frac{1}{\sqrt{2}}
The equality holds at (a,b)=(5/2,2)(a, b) = (5/2, 2).

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