How to put numbers $1$ to $8$ in a row that every number shouldn't be bigger than the sum of numbers next to it? How to put numbers $1$ to $8$ in a row that every number shouldn't be bigger than the sum of numbers next to it?
My attempt: I tried to put numbers from left to right and count the ways but it was to hard and didn't give the answer.Any hints(not answers).

Suppose that the first three numbers satisfy the condition, but the whole seqence does not.
Since $1+2+3+4>8$ we must necessarily have $a_4>a_1+a_2+a_3$.
Case $1: a_4=7$, in this case we must have that $a_1,a_2,a_3$ are a permutation of $1,2,3$. So there are $3!\times 4!$ counterexamples of this form.
Case $2: a_4=8$, In this case we can have that $a_1,a_2,a_3$ are a permutation of $1,2,3$ or $1,2,4$. In total there are $2\times 3!\times 4!$ counterexamples.

So how many permutations satisfy that $a_1>a_2$ and $a_1+a_2>a_3$. Since clearly the two conditions are independent we must only count the triples $(a_1,a_2,a_3)$ with $a_1>a_2$ with $a_1+a_2<a_3$.
They can be enumerated rather quickly:
$(1,a,k)$ gives $5+4+3+2+1=15$ solutions.
$(2,a,k)$ gives $3+2+1=6$ solutions.
$(3,a,k)$ gives $1$ solution.
So there are $22$ solutions.
This means that there are $\frac{8\times 7 \times 6 - 2\times 22}{2}$ triples that satisfy $a_1<a_2$ and $a_1+a_2\geq a_3$. So our final answer is: 
$$\frac{8\times 7 \times 6 - 2\times 22}{2}\times 5!-3\times 3!\times 4!$$
