Prove inequality $|a-b+c-d| \leqslant \frac{1}{16}$

Question

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.

Answer 1

Your conclusion is correct. One can shorten the argument as follows:

$$ (a-b+c-d)^2 = (a+b+c+d)^2 - 4(a+c)(b+d) \\ = (a+b+c+d)^2 - 4(ab+bc+cd+da)\le 4 - 4 = 0 $$ which implies $a-b+c-d = 0$.

Answer 2

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 $$|a-b+c-d|=0<\frac{1}{16}.$$