<s>[INST] Help with this inequality I am given four numbers $a,b,c,d$, such that $c>a>b,c>d>b$ and $0 \le a,b,c,d\le 1$ Can the following two inequalities hold strictly
$ad\le bc$ and $(1-a)(1-d)\le (1-b)(1-c)$.
 [/INST] My interpretation of your question is that you want to find a set of 4 real numbers such that $0 \leq a, b, c, d \leq 1$, $c > d > b$, $c > a > b$, $ad < bc$, $(1-a)(1-d) < (1-b)(1-c)$
If so, then no set of numbers exist.
Consider $[\frac{1}{c^2} - \frac{1}{c} ] (bc) + [\frac{1}{c}] ( 1-b)(1-c) + [ 1 - \frac{1}{c}] \\= \frac{1}{c^2} (bc) - \frac{1}{c} b - \frac{1}{c} c + 1 = 0 $
Note that the coefficients of $(bc)$ and $(1-b)(1-c)$ are positive. Hence, applying the inequalities, we get that 
$ 0 > [\frac{1}{c^2} - \frac{1}{c} ] (ad) + [\frac{1}{c}] (1-a)(1-d) + [ 1 - \frac{1}{c}] \\= \frac{1}{c^2}(ad) - \frac{1}{c} a - \frac{1}{c} d + 1 \\= ( \frac{a}{c} -1) ( \frac{d}{c} -1 ) $
However, the terms on the RHS are both negative, which is a contradiction.
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