AUT_ABooklet_2023

Question

Let a, b, c, d be real numbers with 0 < a, b, c, d < 1 and a + b + c + d = 2. Show that (1 - a)(1 - b)(1 - c)(1 - d) ac + bd2. Are there infinitely many cases of equality?

Accepted Answer

Squaring the given inequality and multiplying by

16

, we get

(2 - 2 a) (2 - 2 b) (2 - 2 c) (2 - 2 d) \leq 4 (a c + b d)^{2} .

We homogenize by replacing the first

2

in each parenthesis on the left side by

a + b + c + d

and get the homogeneous inequality

(b + d - (a - c)) (a + c - (b - d)) (b + d + a - c) (a + c + b - d) \leq 4 (a c + b d)^{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}) = (2 a c + 2 b d + a^{2} + c^{2} - b^{2} - d^{2}) (2 a c + 2 b d - a^{2} - c^{2} + b^{2} + d^{2}) = 4 (a c + b d)^{2} - (a^{2} + c^{2} - b^{2} - d^{2})^{2} \leq 4 (a c + b d)^{2},

which proves the inequality.

Equality holds for

a^{2} + c^{2} = b^{2} + d^{2}

, in particular for

a = b

and

c = d = 1 - a

with

0 < a < 1

. Therefore, there are infinitely many equality cases.