Question 1016849
Let {{{s[1] = 17}}}, and {{{s[2] = 2s[3]-1}}}

Case 1: {{{s[1] + s[2] > s[3]}}} doesn't give any insight, since it reduces to {{{s[3] > -16}}}, which is always true.

Case 2: {{{s[1] + s[3] > s[2]}}} will lead, upon the proper substitution, to {{{17 + s[3] > 2s[3]-1}}}, or {{{18 > s[3]}}}

Case 3:  {{{s[3] + s[2] > s[1]}}} leads to {{{2s[3]-1+s[3] >17}}}, or 
{{{3s[3] > 18}}}, or {{{s[3] > 6}}}

Hence, {{{18 > s[3] > 6}}}.

Now {{{s[3] = (s[2]+1)/2}}}, and so substitution into the last inequality and simplifying yields

{{{35 > s[2] > 11}}}.