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Algebra Difficulty 6.4 National olympiad Prove it Austria

Let aa, bb, cc, dd be real numbers with 0<a,b,c,d<10 < a, b, c, d < 1 and a+b+c+d=2a + b + c + d = 2. Show that
(1−a)(1−b)(1−c)(1−d)≤ac+bd2. \sqrt{(1 - a)(1 - b)(1 - c)(1 - d)} \le \frac{ac + bd}{2}.
Are there infinitely many cases of equality?

Solution

Squaring the given inequality and multiplying by 1616, we get
(2−2a)(2−2b)(2−2c)(2−2d)≤4(ac+bd)2. (2 - 2a)(2 - 2b)(2 - 2c)(2 - 2d) \le 4(ac + bd)^2.
We homogenize by replacing the first 22 in each parenthesis on the left side by a+b+c+da + b + c + d and get the homogeneous inequality
(b+d−(a−c))(a+c−(b−d))(b+d+a−c)(a+c+b−d)≤4(ac+bd)2. (b + d - (a - c))(a + c - (b - d))(b + d + a - c)(a + c + b - d) \le 4(ac + bd)^2.
We evaluate the left-hand side by repeatedly combining two factors and get
(b+d−(a−c))(a+c−(b−d))(b+d+a−c)(a+c+b−d)=((a+c)2−(b−d)2)((b+d)2−(a−c)2)=(2ac+2bd+a2+c2−b2−d2)(2ac+2bd−a2−c2+b2+d2)=4(ac+bd)2−(a2+c2−b2−d2)2≤4(ac+bd)2, \begin{align*} & (b + d - (a - c))(a + c - (b - d))(b + d + a - c)(a + c + b - d) \\ &= ((a + c)^2 - (b - d)^2)((b + d)^2 - (a - c)^2) \\ &= (2ac + 2bd + a^2 + c^2 - b^2 - d^2)(2ac + 2bd - a^2 - c^2 + b^2 + d^2) \\ &= 4(ac + bd)^2 - (a^2 + c^2 - b^2 - d^2)^2 \le 4(ac + bd)^2, \end{align*}
which proves the inequality.

Equality holds for a2+c2=b2+d2a^2 + c^2 = b^2 + d^2, in particular for a=ba = b and c=d=1−ac = d = 1 - a with 0<a<10 < a < 1. Therefore, there are infinitely many equality cases.

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