<s>[INST] Proof that inequality $|a| < |b-c+d|, |b| < |a-c+d|, |c| < |a-b+d|, |d|< |a-b+c| $ does not have solution in $\mathbb{R}$ 
Proof that inequality $$|a| < |b-c+d|, \\|b| < |a-c+d|, \\|c| < |a-b+d|,\\ |d|< |a-b+c| $$ does not have solution in $\mathbb{R}$. 

Solution:
$$|a| < |b-c+d| \leq |b|+|c|+|d|$$
$$|b| < |a-c+d| \leq |a|+|c|+|d|$$
$$|c| < |a-b+d| \leq |a|+|b|+|d|$$
$$|d| < |a-c+c| \leq |a|+|b|+|c|$$
and next 
$$ |a| + |b| + |c| + |d| < 3|a| + 3|b| + 3|c| + 3|d|$$
$$ 0 < 2|a| + 2|b| + 2|c| + 2|d|$$
And I have problem, 
but we have any $a,b,c,d > 0$ 
Where is mistake ? 
 [/INST] There is no mistake, but the inequality you obtained is not useful! 
Note that 
$$|a| < |b-c+d|\Leftrightarrow (b−c+d)^2−a^2>0$$
and similarly we have that $(a-c+d)^2-b^2>0$, $(a-b+d)^2-c^2>0$, and $(a-b+c)^2-d^2>0$. Then by taking the their product, and by recalling that
$x^2-y^2=(x+y)(x-y)$, we find
$$\begin{align}((&b-c+d)^2-a^2)((a-c+d)^2-b^2)((a-b+d)^2-c^2)((a-b+c)^2-d^2)\\
&=-(b-c+d+a)^2(b-c+d-a)^2(a-c+d-b)^2(a-b+c+d)^2
\end{align}$$
which is a contradiction because the left-hand side is $>0$, whereas the right-hand side is $\leq 0$.
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