Inequality Chain

From the inequalities, we can form a "chain" of inequalities $b+c<d+e<a+b<c+d<e+a$ .

From the first 3 terms, we can deduce $c<a,e<c, \text{ and } b<e$ by comparing each individual term to the second term to the right of it $(b+c<a+b, \hspace{.3cm} d+e<c+d, \hspace{.3cm} a+b<e+a)$ .

Once again, we can form the "chain" of inequalities $b<e<c<a$ .

Because $e<c$ , and $b+c<d+e$ , then $d$ must be larger than $b$ , else we will get a contradiction.

Similarly, because $e<c$ , and $c+d<e+a$ , then $d$ must be smaller than $a$ , else we will get a contradiction.

From the above 2 statements, we can deduce $b<d<a$ .

Beacause $b<d<a$ , and $b<e<c<a$ , in all possible cases or "chain", $b$ is always the smallest, $a$ is always the greatest.

b and c appeared most frequently on the LHS and least frequently on the RHS. So either b or c is smallest. Similarly, either a or e is the largest. The only answer which fits is (a, b). (This won't work if more options were given but a quick solution for this question nonetheless)

mallest is b and largest is a. Solution: From all the above we get: b + c < d + e < a + b < c + d < e + a. We can say: c < a, b < d, e < c, d < a, b < e. Therefore: b < a, c, d, e which means b is the smallest. And a > b, c, d, e which means a is the largest