2018 Kazakhstan MO #4

 Prove that for all reals $a,b,c,d\in(0,1)$ we have$$\left(ab-cd\right)\left(ac+bd\right)\left(ad-bc\right)+\min{\left(a,b,c,d\right)} < 1.$$



WLOGE assume $ a \ge c. $ Because $ a, b, d <1 $, by Cauchy-Schwarz we get $$(ab-cd)(ad-bc)\le \left(\frac{(ab-cd)+(ad-bc)}{2}\right)^ 2=\frac{(b+d)^2(a-c)^2}{4} < (1-c)^2.$$

Also, $ac + bd <c + 1$. Multiplying our two inequalities, we obtain $$ (ab-cd) (ac + bd) (ad-bc) <(1-c) ^ 2 (1 + c) = ( 1-c) (1-c ^ 2) <1-c$$ $$\le 1- \min (a, b, c, d),$$ as desired. $\square$

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