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Taiwan IMO 1J, 2020

Algebra Difficulty 8.2 Shortlist Prove it Taiwan

Let a,b,c,da, b, c, d be real numbers satisfying
(a+c)(b+d)=2(ac−2bd−1). (a + c)(b + d) = \sqrt{2}(ac - 2bd - 1).
Show that
(ab−1)2+(bc−1)2+(cd−1)2+(da−1)2+(ac−1)2+(2bd+1)2≥4. (ab - 1)^2 + (bc - 1)^2 + (cd - 1)^2 + (da - 1)^2 + (ac - 1)^2 + (2bd + 1)^2 \geq 4.

Solution

Let A=(a+c)(b+d)=2(ac−2bd−1)A = (a+c)(b+d) = \sqrt{2}(ac-2bd-1). Note that
∑cyc(ab−1)2≥∑cyc(ab−1)2−(∑cycab−1)2=−2(∑cycab2c)−4abcd+3=3−4abcd−2(ab+cd)(ad+bc)=3−4abcd−2(ab+cd)(A−ab−cd)=3−4abcd+2(ab+cd−A2)2−A22≥3−4abcd−(ac−2bd−1)2=4−(ac−1)2−(2bd+1)2 \begin{align*} \sum_{\text{cyc}} (ab-1)^2 &\ge \sum_{\text{cyc}} (ab-1)^2 - \left(\sum_{\text{cyc}} ab-1\right)^2 \\ &= -2 \left(\sum_{\text{cyc}} ab^2 c\right) - 4abcd + 3 \\ &= 3 - 4abcd - 2(ab + cd)(ad + bc) \\ &= 3 - 4abcd - 2(ab + cd)(A - ab - cd) \\ &= 3 - 4abcd + 2 \left(ab + cd - \frac{A}{2}\right)^2 - \frac{A^2}{2} \\ &\ge 3 - 4abcd - (ac - 2bd - 1)^2 \\ &= 4 - (ac - 1)^2 - (2bd + 1)^2 \end{align*}
Hence we obtain
(ab−1)2+(bc−1)2+(cd−1)2+(da−1)2+(ac−1)2+(2bd+1)2≥4. (ab - 1)^2 + (bc - 1)^2 + (cd - 1)^2 + (da - 1)^2 + (ac - 1)^2 + (2bd + 1)^2 \geq 4.

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