<s>[INST] Prove inequality $|a-b+c-d| \leqslant \frac{1}{16}$ Let $a,b,c,d$ be positive real numbers that fulfill two conditions:
$$a+b+c+d \leqslant 2$$$$ab+bc+cd+ad \geqslant 1$$
Prove that $|a-b+c-d|\leqslant \frac{1}{16}$
Let:
$a+c=x$ and $b+d=y$
Both $x$ and $y$ are positive.
$$x+y \leqslant 2$$$$xy \geqslant 1$$
$$-4xy \leqslant -4$$
$$(x+y)^2 \leqslant 4$$
$$(x-y)^2 \leqslant 0 \Rightarrow |a-b+c-d|=|x-y|=0\leqslant \frac{1}{16}$$
I do not really think that this is valid solution, but I can not find any mistakes.
 [/INST] By AM-GM $$1\leq ab+bc+cd+da=(a+c)(b+d)\leq\left(\frac{a+c+b+d}{2}\right)^2\leq1,$$
which gives $$a+c=b+d$$ and $$a+b+c+d=2$$ or
$$a+c=b+d=1,$$ which gives $$|a-b+c-d|=0<\frac{1}{16}.$$
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