Alice has $n$ positive real numbers $a_1,a_2,.....,a_n$ such that $a_1 \leq a_2 \leq a_3.......\leq a_n$ and $a_{i+1}\leq 2a_i$ for all $i=1,2,3,4.....,n-1$. Alice suggested Bob to choose signs $'+'$ and $'-'$ such that $k=\pm a_1\pm a_2....\pm a_n$ satisfies $0 \leq k \leq a_1$.
Can Bob always do that ?
I was wondering of how can I approach this problem ? Does anyone have any idea of how can I proceed ? Any help would be appreciated.
Proof by Induction. Trivially true for $n=1$ and for $n=2$ take $a_2-a_1$.
Suppose true for $n-1$. Apply it to the chain $a_2≤\cdots≤a_n$ to get $k_1=\pm a_2+\cdots +\pm a_n$ with $0≤k_1≤a_2$. But then one of $a_1-k_1$ and $k_1-a_1$ will work.